mirror of
https://github.com/massbug/judge4c.git
synced 2025-10-31 15:45:40 +00:00
1111 lines
32 KiB
TypeScript
1111 lines
32 KiB
TypeScript
import { PrismaClient, Prisma, EditorLanguage, LanguageServerProtocol } from "@/generated/client";
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const prisma = new PrismaClient();
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const editorLanguageConfigData: Prisma.EditorLanguageConfigCreateInput[] = [
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{
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language: EditorLanguage.c,
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label: "C",
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fileName: "main",
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fileExtension: ".c",
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languageServerConfig: {
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create: {
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protocol: LanguageServerProtocol.ws,
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hostname: "localhost",
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port: 4594,
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path: "/clangd",
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},
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},
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dockerConfig: {
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create: {
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image: "gcc",
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tag: "latest",
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workingDir: "/src",
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compileOutputLimit: 1 * 1024 * 1024,
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runOutputLimit: 1 * 1024 * 1024,
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},
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},
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},
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{
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language: EditorLanguage.cpp,
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label: "C++",
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fileName: "main",
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fileExtension: ".cpp",
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languageServerConfig: {
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create: {
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protocol: LanguageServerProtocol.ws,
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hostname: "localhost",
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port: 4595,
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path: "/clangd",
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},
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},
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dockerConfig: {
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create: {
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image: "gcc",
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tag: "latest",
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workingDir: "/src",
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compileOutputLimit: 1 * 1024 * 1024,
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runOutputLimit: 1 * 1024 * 1024,
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},
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},
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},
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];
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const userData: Prisma.UserCreateInput[] = [
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{
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name: "cfngc4594",
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email: "cfngc4594@gmail.com",
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password: "$2b$10$edWXpq2TOiiGQkPOXWKGlO4EKnp2YyV7OoS2qqk/W0E6GyiVQIC66",
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role: "ADMIN",
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problems: {
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create: [
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{
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displayId: 1000,
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title: "两数之和",
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description: `给定一个整数数组 \`nums\` 和一个整数目标值 \`target\`，请你在该数组中找出 **和为目标值** _\`target\`_ 的那 **两个** 整数，并返回它们的数组下标。
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你可以假设每种输入只会对应一个答案，并且你不能使用两次相同的元素。
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你可以按任意顺序返回答案。
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**示例 1：**
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\`\`\`shell
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输入：nums = [2,7,11,15], target = 9
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输出：[0,1]
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解释：因为 nums[0] + nums[1] == 9 ，返回 [0, 1] 。
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\`\`\`
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### 示例 2：
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\`\`\`shell
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输入：nums = [3,2,4], target = 6
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输出：[1,2]
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\`\`\`
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###示例 3：
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\`\`\`shell
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输入：nums = [3,3], target = 6
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输出：[0,1]
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\`\`\`
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##提示：
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\`\`\`math
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2 <= nums.length <= 10^4
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\`\`\`
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\`\`\`math
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-10^9 <= nums[i] <= 10^9
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\`\`\`
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\`\`\`math
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-10^9 <= target <= 10^9
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\`\`\`
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<div align="center">
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只会存在一个有效答案*
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</div>
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**进阶：** 你可以想出一个时间复杂度小于 \`O(n^2)\` 的算法吗？
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---
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\`\`\`C++
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\`\`\``,
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solution: `<VideoEmbed platform="youtube" id="tSI98g3PDyE" />
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## 方法一：暴力枚举
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### 思路及算法
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最容易想到的方法是枚举数组中的每一个数 x，寻找数组中是否存在 target - x。
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当我们使用遍历整个数组的方式寻找 target - x 时，需要注意到每一个位于 x 之前的元素都已经和 x 匹配过，因此不需要再进行匹配。而每一个元素不能被使用两次，所以我们只需要在 x 后面的元素中寻找 target - x。
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### 代码
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\`\`\`c showLineNumbers
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int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
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struct hashTable {
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int key;
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int value;
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UT_hash_handle hh;
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} *hashTable = NULL, *item, *tmpItem;
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for (int i = 0; i < numsSize; i++) {
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HASH_FIND_INT(hashTable, &nums[i], item);
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if (item) {
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int* result = malloc(sizeof(int) * 2);
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result[0] = item->value;
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result[1] = i;
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*returnSize = 2;
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HASH_ITER(hh, hashTable, item, tmpItem) {
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HASH_DEL(hashTable, item);
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free(item);
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}
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return result;
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}
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item = malloc(sizeof(struct hashTable));
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item->key = target - nums[i];
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item->value = i;
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HASH_ADD_INT(hashTable, key, item);
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}
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HASH_ITER(hh, hashTable, item, tmpItem) {
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HASH_DEL(hashTable, item);
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free(item);
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}
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*returnSize = 0;
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// If no valid pair is found, return an empty array
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return malloc(sizeof(int) * 0);
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}
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\`\`\`
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### 复杂度分析
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- **时间复杂度:** $O(n^2)$.
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其中 $n$ 是数组中的元素数量。最坏情况下数组中任意两个数都要被匹配一次。
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- **空间复杂度:** $O(1)$.
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所需的空间不取决于 input 数组的大小，因此仅使用恒定空间。
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---
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## 方法二：哈希表
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### 思路及算法
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注意到方法一的时间复杂度较高的原因是寻找 target - x 的时间复杂度过高。因此，我们需要一种更优秀的方法，能够快速寻找数组中是否存在目标元素。如果存在，我们需要找出它的索引。
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使用哈希表，可以将寻找 target - x 的时间复杂度降低到从 $O(N)$ 降低到 $O(1)$。
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这样我们创建一个哈希表，对于每一个 x，我们首先查询哈希表中是否存在 target - x，然后将 x 插入到哈希表中，即可保证不会让 x 和自己匹配。
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### 代码
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\`\`\`c showLineNumbers
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int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
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struct hashTable {
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int key;
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int value;
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UT_hash_handle hh;
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} *hashTable = NULL, *item, *tmpItem;
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for (int i = 0; i < numsSize; i++) {
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HASH_FIND_INT(hashTable, &nums[i], item);
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if (item) {
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int* result = malloc(sizeof(int) * 2);
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result[0] = item->value;
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result[1] = i;
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*returnSize = 2;
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HASH_ITER(hh, hashTable, item, tmpItem) {
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HASH_DEL(hashTable, item);
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free(item);
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}
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return result;
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}
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item = malloc(sizeof(struct hashTable));
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item->key = target - nums[i];
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item->value = i;
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HASH_ADD_INT(hashTable, key, item);
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}
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HASH_ITER(hh, hashTable, item, tmpItem) {
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HASH_DEL(hashTable, item);
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free(item);
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}
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*returnSize = 0;
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// If no valid pair is found, return an empty array
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return malloc(sizeof(int) * 0);
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}
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\`\`\`
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### 复杂度分析
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- **时间复杂度:** $O(n)$.
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其中 $N$ 是数组中的元素数量。对于每一个元素 x，我们可以 $O(1)$ 地寻找 target - x。
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- **空间复杂度:** $O(n)$.
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其中 $N$ 是数组中的元素数量。主要为哈希表的开销。
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`,
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difficulty: "EASY",
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published: true,
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templates: {
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create: [
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{
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language: "c",
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template: `
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#include <stdio.h>
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#include <stdlib.h>
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#include <string.h>
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int *parseIntArray(char *line, int *len) {
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line[strcspn(line, "\\n")] = 0;
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char *p = line;
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while (*p && (*p == '[' || *p == ' ' || *p == ']'))
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p++;
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int capacity = 10;
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int *arr = malloc(capacity * sizeof(int));
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*len = 0;
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char *token = strtok(p, ",");
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while (token) {
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if (*len >= capacity) {
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capacity *= 2;
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arr = realloc(arr, capacity * sizeof(int));
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}
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arr[(*len)++] = atoi(token);
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token = strtok(NULL, ",");
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}
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return arr;
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}
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char *formatOutput(int *res, int resLen) {
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if (resLen == 0)
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return "[]";
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char *buf = malloc(resLen * 12 + 3);
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char *p = buf;
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*p++ = '[';
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for (int i = 0; i < resLen; i++) {
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p += sprintf(p, "%d", res[i]);
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if (i != resLen - 1)
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*p++ = ',';
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}
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*p++ = ']';
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*p = 0;
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return buf;
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}
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int *twoSum(int *nums, int numsSize, int target, int *returnSize);
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int main() {
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char line[1024];
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while (fgets(line, sizeof(line), stdin)) {
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int numsSize;
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int *nums = parseIntArray(line, &numsSize);
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if (!fgets(line, sizeof(line), stdin))
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break;
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int target = atoi(line);
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int returnSize;
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int *res = twoSum(nums, numsSize, target, &returnSize);
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char *output = formatOutput(res, returnSize);
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printf("%s\\n", output);
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free(nums);
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if (returnSize > 0)
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free(res);
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if (returnSize > 0)
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free(output);
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}
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return 0;
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}
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/**
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* Note: The returned array must be malloced, assume caller calls free().
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*/
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int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
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}`,
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},
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{
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language: "cpp",
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template: `
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#include <iostream>
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#include <vector>
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#include <string>
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#include <sstream>
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#include <algorithm>
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#include <unordered_map>
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using namespace std;
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// 解析输入字符串为整数数组
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vector<int> parseIntArray(string line) {
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vector<int> result;
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line.erase(remove(line.begin(), line.end(), '['), line.end());
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line.erase(remove(line.begin(), line.end(), ']'), line.end());
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stringstream ss(line);
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string token;
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while (getline(ss, token, ',')) {
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if (!token.empty()) {
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result.push_back(stoi(token));
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}
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}
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return result;
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}
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// 格式化输出结果为字符串
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string formatOutput(const vector<int>& res) {
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if (res.empty()) return "[]";
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stringstream ss;
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ss << "[";
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for (size_t i = 0; i < res.size(); ++i) {
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ss << res[i];
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if (i != res.size() - 1)
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ss << ",";
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}
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ss << "]";
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return ss.str();
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}
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// Solution 类声明
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class Solution {
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public:
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vector<int> twoSum(vector<int>& nums, int target);
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};
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int main() {
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string line;
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while (getline(cin, line)) {
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vector<int> nums = parseIntArray(line);
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if (!getline(cin, line)) break;
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int target = stoi(line);
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Solution sol;
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vector<int> res = sol.twoSum(nums, target);
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cout << formatOutput(res) << endl;
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}
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return 0;
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}
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vector<int> Solution::twoSum(vector<int>& nums, int target) {
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return {};
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}
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`,
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},
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],
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},
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testcases: {
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create: [
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{
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data: {
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create: [
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{ label: "nums", value: "[2,7,11,15]", index: 0 },
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{ label: "target", value: "9", index: 1 },
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],
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},
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expectedOutput: "[0,1]",
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},
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{
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data: {
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create: [
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{ label: "nums", value: "[3,2,4]", index: 0 },
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{ label: "target", value: "6", index: 1 },
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],
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},
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expectedOutput: "[1,2]",
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},
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{
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data: {
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create: [
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{ label: "nums", value: "[3,3]", index: 0 },
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{ label: "target", value: "6", index: 1 },
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],
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},
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expectedOutput: "[0,1]",
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},
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],
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},
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},
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{
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displayId: 1001,
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title: "两数相加",
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description: `给你两个 **非空** 的链表，表示两个非负的整数。它们每位数字都是按照 **逆序** 的方式存储的，并且每个节点只能存储 **一位** 数字。
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请你将两个数相加，并以相同形式返回一个表示和的链表。
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你可以假设除了数字 0 之外，这两个数都不会以 0 开头。
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###示例 1：
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\`\`\`shell
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输入：l1 = [2,4,3], l2 = [5,6,4]
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输出：[7,0,8]
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解释：342 + 465 = 807.
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\`\`\`
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**示例 2：**
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\`\`\`shell
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输入：l1 = [0], l2 = [0]
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输出：[0]
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\`\`\`
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###示例 3：
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\`\`\`shell
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输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
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输出：[8,9,9,9,0,0,0,1]
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\`\`\`
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## 提示：
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<div align="center">
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每个链表中的节点数在范围 $[1, 100]$ 内
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\`0 <= Node.val <= 9\`
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题目数据保证列表表示的数字不含前导零
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</div>
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---
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\`\`\`C++
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||
\`\`\``,
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solution: `## 方法一：模拟
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### 思路与算法
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||
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由于输入的两个链表都是逆序存储数字的位数的，因此两个链表中同一位置的数字可以直接相加。
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我们同时遍历两个链表，逐位计算它们的和，并与当前位置的进位值相加。具体而言，如果当前两个链表处相应位置的数字为 n1,n2，进位值为 carry，则它们的和为 n1+n2+carry；其中，答案链表处相应位置的数字为 (n1+n2+carry)mod10，而新的进位值为 ⌊
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10
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如果两个链表的长度不同，则可以认为长度短的链表的后面有若干个 $0$ 。
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此外，如果链表遍历结束后，有 $carry>0$，还需要在答案链表的后面附加一个节点，节点的值为 carry。
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### 代码
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||
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\`\`\`c showLineNumbers
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struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
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struct ListNode* dummyHead = malloc(sizeof(struct ListNode));
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dummyHead->val = 0;
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dummyHead->next = NULL;
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struct ListNode* curr = dummyHead;
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int carry = 0;
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while (l1 != NULL || l2 != NULL || carry != 0) {
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int x = (l1 != NULL) ? l1->val : 0;
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int y = (l2 != NULL) ? l2->val : 0;
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int sum = carry + x + y;
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carry = sum / 10;
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curr->next = malloc(sizeof(struct ListNode));
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curr->next->val = sum % 10;
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curr->next->next = NULL;
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curr = curr->next;
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if (l1 != NULL) l1 = l1->next;
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if (l2 != NULL) l2 = l2->next;
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}
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struct ListNode* result = dummyHead->next;
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free(dummyHead); // Free the memory allocated for dummyHead
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return result;
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}
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\`\`\`
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||
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||
### 复杂度分析
|
||
|
||
- **时间复杂度:** $O(max(m,n))$
|
||
|
||
其中 $m$ 和 $n$ 分别为两个链表的长度。我们要遍历两个链表的全部位置，而处理每个位置只需要 $O(1)$ 的时间。
|
||
|
||
- **空间复杂度:** $O(1)$
|
||
|
||
注意返回值不计入空间复杂度。
|
||
|
||
`,
|
||
difficulty: "MEDIUM",
|
||
published: true,
|
||
templates: {
|
||
create: [
|
||
{
|
||
language: "c",
|
||
template: `
|
||
#include <stdio.h>
|
||
#include <stdlib.h>
|
||
#include <string.h>
|
||
|
||
// Definition for singly-linked list.
|
||
struct ListNode {
|
||
int val;
|
||
struct ListNode *next;
|
||
};
|
||
|
||
// 创建链表
|
||
struct ListNode* createList(char *line) {
|
||
struct ListNode dummy;
|
||
struct ListNode *tail = &dummy;
|
||
dummy.next = NULL;
|
||
|
||
line[strcspn(line, "\\n")] = 0;
|
||
char *p = line;
|
||
while (*p && (*p == '[' || *p == ' ' || *p == ']')) p++;
|
||
|
||
char *token = strtok(p, ",");
|
||
while (token) {
|
||
struct ListNode *node = malloc(sizeof(struct ListNode));
|
||
node->val = atoi(token);
|
||
node->next = NULL;
|
||
tail->next = node;
|
||
tail = node;
|
||
token = strtok(NULL, ",");
|
||
}
|
||
|
||
return dummy.next;
|
||
}
|
||
|
||
// 打印链表
|
||
void printList(struct ListNode* head) {
|
||
printf("[");
|
||
while (head) {
|
||
printf("%d", head->val);
|
||
if (head->next) printf(",");
|
||
head = head->next;
|
||
}
|
||
printf("]\\n");
|
||
}
|
||
|
||
// 释放链表内存
|
||
void freeList(struct ListNode* head) {
|
||
while (head) {
|
||
struct ListNode* temp = head;
|
||
head = head->next;
|
||
free(temp);
|
||
}
|
||
}
|
||
|
||
// 主函数
|
||
int main() {
|
||
char line[1024];
|
||
|
||
while (fgets(line, sizeof(line), stdin)) {
|
||
struct ListNode* l1 = createList(line);
|
||
|
||
if (!fgets(line, sizeof(line), stdin)) break;
|
||
struct ListNode* l2 = createList(line);
|
||
|
||
struct ListNode* result = addTwoNumbers(l1, l2);
|
||
printList(result);
|
||
|
||
freeList(l1);
|
||
freeList(l2);
|
||
freeList(result);
|
||
}
|
||
|
||
return 0;
|
||
}
|
||
|
||
|
||
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
|
||
|
||
return NULL; // 在这里填充你的算法逻辑
|
||
}
|
||
|
||
`,
|
||
},
|
||
{
|
||
language: "cpp",
|
||
template: `
|
||
#include <iostream>
|
||
#include <string>
|
||
#include <sstream>
|
||
#include <vector>
|
||
#include <algorithm>
|
||
using namespace std;
|
||
|
||
// Definition for singly-linked list.
|
||
struct ListNode {
|
||
int val;
|
||
ListNode *next;
|
||
ListNode() : val(0), next(nullptr) {}
|
||
ListNode(int x) : val(x), next(nullptr) {}
|
||
ListNode(int x, ListNode *next) : val(x), next(next) {}
|
||
};
|
||
|
||
// 声明 Solution 类
|
||
class Solution {
|
||
public:
|
||
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2);
|
||
};
|
||
|
||
// 输入字符串 -> 链表
|
||
ListNode* createList(const string& line) {
|
||
ListNode dummy;
|
||
ListNode* tail = &dummy;
|
||
dummy.next = nullptr;
|
||
|
||
string nums = line;
|
||
nums.erase(remove(nums.begin(), nums.end(), '['), nums.end());
|
||
nums.erase(remove(nums.begin(), nums.end(), ']'), nums.end());
|
||
|
||
stringstream ss(nums);
|
||
string token;
|
||
while (getline(ss, token, ',')) {
|
||
if (!token.empty()) {
|
||
int val = stoi(token);
|
||
tail->next = new ListNode(val);
|
||
tail = tail->next;
|
||
}
|
||
}
|
||
|
||
return dummy.next;
|
||
}
|
||
|
||
// 打印链表
|
||
void printList(ListNode* head) {
|
||
cout << "[";
|
||
while (head) {
|
||
cout << head->val;
|
||
if (head->next) cout << ",";
|
||
head = head->next;
|
||
}
|
||
cout << "]" << endl;
|
||
}
|
||
|
||
// 释放内存
|
||
void freeList(ListNode* head) {
|
||
while (head) {
|
||
ListNode* tmp = head;
|
||
head = head->next;
|
||
delete tmp;
|
||
}
|
||
}
|
||
|
||
// 主函数
|
||
int main() {
|
||
string line;
|
||
while (getline(cin, line)) {
|
||
ListNode* l1 = createList(line);
|
||
if (!getline(cin, line)) break;
|
||
ListNode* l2 = createList(line);
|
||
|
||
Solution sol;
|
||
ListNode* res = sol.addTwoNumbers(l1, l2);
|
||
printList(res);
|
||
|
||
freeList(l1);
|
||
freeList(l2);
|
||
freeList(res);
|
||
}
|
||
return 0;
|
||
}
|
||
|
||
|
||
ListNode* Solution::addTwoNumbers(ListNode* l1, ListNode* l2) {
|
||
|
||
return nullptr; // 在这里填充你的算法逻辑
|
||
}
|
||
`,
|
||
},
|
||
],
|
||
},
|
||
testcases: {
|
||
create: [
|
||
{
|
||
data: {
|
||
create: [
|
||
{ label: "l1", value: "[2,4,3]", index: 0 },
|
||
{ label: "l2", value: "[5,6,4]", index: 1 },
|
||
],
|
||
},
|
||
expectedOutput: "[7,0,8]",
|
||
},
|
||
{
|
||
data: {
|
||
create: [
|
||
{ label: "l1", value: "[0]", index: 0 },
|
||
{ label: "l2", value: "[0]", index: 1 },
|
||
],
|
||
},
|
||
expectedOutput: "[0]",
|
||
},
|
||
{
|
||
data: {
|
||
create: [
|
||
{ label: "l1", value: "[9,9,9,9,9,9,9]", index: 0 },
|
||
{ label: "l2", value: "[9,9,9,9]", index: 1 },
|
||
],
|
||
},
|
||
expectedOutput: "[8,9,9,9,0,0,0,1]",
|
||
},
|
||
],
|
||
},
|
||
},
|
||
],
|
||
},
|
||
},
|
||
{
|
||
name: "fly6516",
|
||
email: "fly6516@outlook.com",
|
||
password: "$2b$10$SD1T/dYvKTArGdTmf8ERxuBKIONxY01/wSboRNaNsHnKZzDhps/0u",
|
||
role: "ADMIN",
|
||
problems: {
|
||
create: [
|
||
{
|
||
displayId: 1002,
|
||
title: "寻找两个正序数组的中位数",
|
||
description: `给定两个大小分别为 \`m\` 和 \`n\` 的正序（从小到大）数组 \`nums1\` 和 \`nums2\`。请你找出并返回这两个正序数组的 **中位数** 。
|
||
|
||
算法的时间复杂度应该为 \`O(log (m+n))\` 。
|
||
|
||
### 示例 1：
|
||
|
||
\`\`\`shell
|
||
输入：nums1 = [1,3], nums2 = [2]
|
||
输出：2.00000
|
||
解释：合并数组 = [1,2,3] ，中位数 2
|
||
\`\`\`
|
||
|
||
### 示例 2：
|
||
|
||
\`\`\`shell
|
||
输入：nums1 = [1,2], nums2 = [3,4]
|
||
输出：2.50000
|
||
解释：合并数组 = [1,2,3,4] ，中位数 (2 + 3) / 2 = 2.5
|
||
\`\`\`
|
||
|
||
## 提示：
|
||
|
||
\`\`\`math
|
||
nums1.length == m
|
||
\`\`\`
|
||
\`\`\`math
|
||
nums2.length == n
|
||
\`\`\`
|
||
\`\`\`math
|
||
0 <= m <= 1000
|
||
\`\`\`
|
||
\`\`\`math
|
||
0 <= n <= 1000
|
||
\`\`\`
|
||
\`\`\`math
|
||
1 <= m + n <= 2000
|
||
\`\`\`
|
||
\`\`\`math
|
||
-10^6 <= nums1[i], nums2[i] <= 10^6
|
||
\`\`\`
|
||
|
||
---
|
||
|
||
|
||
|
||
\`\`\`C++
|
||
\`\`\``,
|
||
solution: `## 方法一：二分查找
|
||
|
||
### Intuition
|
||
|
||
给定两个有序数组，要求找到两个有序数组的中位数，最直观的思路有以下两种：
|
||
|
||
- 使用归并的方式，合并两个有序数组，得到一个大的有序数组。大的有序数组的中间位置的元素，即为中位数。
|
||
|
||
- 不需要合并两个有序数组，只要找到中位数的位置即可。由于两个数组的长度已知，因此中位数对应的两个数组的下标之和也是已知的。维护两个指针，初始时分别指向两个数组的下标 0 的位置，每次将指向较小值的指针后移一位（如果一个指针已经到达数组末尾，则只需要移动另一个数组的指针），直到到达中位数的位置。
|
||
|
||
|
||
#假设两个有序数组的长度分别为 $m$ 和 $n$，上述两种思路的复杂度如何？
|
||
|
||
第一种思路的时间复杂度是 $O(m+n)$，空间复杂度是 $O(m+n)$。第二种思路虽然可以将空间复杂度降到 $O(1)$，但是时间复杂度仍是 $O(m+n)$。
|
||
|
||
如何把时间复杂度降低到 $O(log(m+n))$ 呢？如果对时间复杂度的要求有 $log$，通常都需要用到二分查找，这道题也可以通过二分查找实现。
|
||
|
||
根据中位数的定义，当 $m+n$ 是奇数时，中位数是两个有序数组中的第 $(m+n)/2$ 个元素，当 $m+n$ 是偶数时，中位数是两个有序数组中的第 $(m+n)/2$ 个元素和第 $(m+n)/2+1$ 个元素的平均值。因此，这道题可以转化成寻找两个有序数组中的第 $k$ 小的数，其中 $k$ 为 $(m+n)/2$ 或 $(m+n)/2+1$。
|
||
|
||
假设两个有序数组分别是 $A$ 和 $B$。要找到第 $k$ 个元素，我们可以比较 $A[k/2−1]$ 和 $B[k/2−1]$，其中 $/$ 表示整数除法。由于 $A[k/2−1]$ 和 $B[k/2−1]$ 的前面分别有 $A[0..k/2−2]$ 和 $B[0..k/2−2]$，即 $k/2−1$ 个元素，对于 $A[k/2−1]$ 和 $B[k/2−1]$ 中的较小值，最多只会有 $(k/2−1)+(k/2−1)≤k−2$ 个元素比它小，那么它就不能是第 $k$ 小的数了。
|
||
|
||
|
||
 {
|
||
// If the segment of on array is empty, it means we have passed all
|
||
// its element, just return the corresponding element in the other array.
|
||
if (aEnd < aStart) {
|
||
return B[k - aStart];
|
||
}
|
||
if (bEnd < bStart) {
|
||
return A[k - bStart];
|
||
}
|
||
|
||
// Get the middle indexes and middle values of A and B.
|
||
int aIndex = (aStart + aEnd) / 2, bIndex = (bStart + bEnd) / 2;
|
||
int aValue = A[aIndex], bValue = B[bIndex];
|
||
|
||
// If k is in the right half of A + B, remove the smaller left half.
|
||
if (aIndex + bIndex < k) {
|
||
if (aValue > bValue) {
|
||
return solve(A, aStart, aEnd, B, bIndex + 1, bEnd, k);
|
||
} else {
|
||
return solve(A, aIndex + 1, aEnd, B, bStart, bEnd, k);
|
||
}
|
||
}
|
||
// Otherwise, remove the larger right half.
|
||
else {
|
||
if (aValue > bValue) {
|
||
return solve(A, aStart, aIndex - 1, B, bStart, bEnd, k);
|
||
} else {
|
||
return solve(A, aStart, aEnd, B, bStart, bIndex - 1, k);
|
||
}
|
||
}
|
||
}
|
||
|
||
double findMedianSortedArrays(int* A, int na, int* B, int nb) {
|
||
int n = na + nb;
|
||
if (n % 2 == 1) {
|
||
return solve(A, 0, na - 1, B, 0, nb - 1, n / 2);
|
||
} else {
|
||
return (solve(A, 0, na - 1, B, 0, nb - 1, n / 2) +
|
||
solve(A, 0, na - 1, B, 0, nb - 1, n / 2 - 1)) /
|
||
2.0;
|
||
}
|
||
}
|
||
\`\`\`
|
||
|
||
### 复杂度分析
|
||
|
||
Let $m$ be the size of array \`nums1\` and $n$ be the size of array \`nums2\`.
|
||
|
||
- **时间复杂度:** $O(log(m+n))$
|
||
|
||
- 其中 m 和 n 分别是数组 nums 1和 nums 2的长度。初始时有 $k=(m+n)/2$ 或 $k=(m+n)/2+1$，每一轮循环可以将查找范围减少一半，因此时间复杂度是 $O(log(m+n))$。
|
||
|
||
- **空间复杂度:** $O(1)$
|
||
|
||
`,
|
||
difficulty: "HARD",
|
||
published: true,
|
||
templates: {
|
||
create: [
|
||
{
|
||
language: "c",
|
||
template: `
|
||
#include <stdio.h>
|
||
#include <stdlib.h>
|
||
#include <string.h>
|
||
|
||
// 解析输入数组
|
||
int *parseIntArray(char *line, int *len) {
|
||
line[strcspn(line, "\\n")] = 0; // 移除换行符
|
||
char *p = line;
|
||
while (*p && (*p == '[' || *p == ' ' || *p == ']')) p++; // 跳过空格和括号
|
||
|
||
int capacity = 10;
|
||
int *arr = malloc(capacity * sizeof(int)); // 初始分配空间
|
||
*len = 0;
|
||
|
||
char *token = strtok(p, ","); // 分割输入为逗号分隔的整数
|
||
while (token) {
|
||
if (*len >= capacity) { // 扩展数组大小
|
||
capacity *= 2;
|
||
arr = realloc(arr, capacity * sizeof(int));
|
||
}
|
||
arr[(*len)++] = atoi(token); // 存储整数
|
||
token = strtok(NULL, ",");
|
||
}
|
||
return arr;
|
||
}
|
||
|
||
double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size);
|
||
|
||
int main() {
|
||
char line[1024];
|
||
|
||
while (fgets(line, sizeof(line), stdin)) { // 读取第一行
|
||
int len1;
|
||
int *nums1 = parseIntArray(line, &len1); // 解析数组1
|
||
|
||
if (!fgets(line, sizeof(line), stdin)) break; // 如果第二行不存在，退出
|
||
int len2;
|
||
int *nums2 = parseIntArray(line, &len2); // 解析数组2
|
||
|
||
double result = findMedianSortedArrays(nums1, len1, nums2, len2); // 计算中位数
|
||
printf("%.5f\\n", result); // 输出中位数，保留5位小数
|
||
|
||
free(nums1); // 释放内存
|
||
free(nums2); // 释放内存
|
||
}
|
||
|
||
return 0;
|
||
}
|
||
|
||
|
||
// 寻找中位数函数
|
||
double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
|
||
|
||
return 0.0; // 在这里填充你的算法逻辑
|
||
}
|
||
`,
|
||
},
|
||
{
|
||
language: "cpp",
|
||
template: `
|
||
#include <iostream>
|
||
#include <vector>
|
||
#include <string>
|
||
#include <sstream>
|
||
#include <algorithm>
|
||
using namespace std;
|
||
|
||
class Solution {
|
||
public:
|
||
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2);
|
||
};
|
||
|
||
// 解析输入为整数数组
|
||
vector<int> parseIntArray(const string& line) {
|
||
string trimmed = line;
|
||
trimmed.erase(remove(trimmed.begin(), trimmed.end(), '['), trimmed.end());
|
||
trimmed.erase(remove(trimmed.begin(), trimmed.end(), ']'), trimmed.end());
|
||
|
||
vector<int> result;
|
||
stringstream ss(trimmed);
|
||
string token;
|
||
while (getline(ss, token, ',')) {
|
||
if (!token.empty()) {
|
||
result.push_back(stoi(token));
|
||
}
|
||
}
|
||
return result;
|
||
}
|
||
|
||
int main() {
|
||
string line;
|
||
while (getline(cin, line)) {
|
||
vector<int> nums1 = parseIntArray(line);
|
||
if (!getline(cin, line)) break;
|
||
vector<int> nums2 = parseIntArray(line);
|
||
|
||
Solution sol;
|
||
double result = sol.findMedianSortedArrays(nums1, nums2);
|
||
printf("%.5f\\n", result);
|
||
}
|
||
return 0;
|
||
}
|
||
|
||
|
||
|
||
double Solution::findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
|
||
|
||
return 0.0; // 临时返回值，待填充
|
||
}
|
||
`,
|
||
},
|
||
],
|
||
},
|
||
testcases: {
|
||
create: [
|
||
{
|
||
data: {
|
||
create: [
|
||
{ label: "nums1", value: "[1,3]", index: 0 },
|
||
{ label: "nums2", value: "[2]", index: 1 },
|
||
],
|
||
},
|
||
expectedOutput: "2.00000",
|
||
},
|
||
{
|
||
data: {
|
||
create: [
|
||
{ label: "nums1", value: "[1,2]", index: 0 },
|
||
{ label: "nums2", value: "[3,4]", index: 1 },
|
||
],
|
||
},
|
||
expectedOutput: "2.50000",
|
||
},
|
||
],
|
||
},
|
||
},
|
||
],
|
||
},
|
||
},
|
||
];
|
||
|
||
export async function main() {
|
||
for (const e of editorLanguageConfigData) {
|
||
await prisma.editorLanguageConfig.create({ data: e });
|
||
}
|
||
|
||
for (const u of userData) {
|
||
await prisma.user.create({ data: u });
|
||
}
|
||
}
|
||
|
||
main();
|