chore(seed): add seed data for problem templates

prisma/seed.ts

@@ -297,11 +297,17 @@ int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
             create: [
               {
                 language: "c",
-                template: `
-#include <stdio.h>
+                template: `#include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
 
+/**
+ * Note: The returned array must be malloced, assume caller calls free().
+ */
+int *twoSum(int *nums, int numsSize, int target, int *returnSize) {
+  
+}
+
 int *parseIntArray(char *line, int *len) {
   line[strcspn(line, "\\n")] = 0;
   char *p = line;
@@ -341,8 +347,6 @@ char *formatOutput(int *res, int resLen) {
   return buf;
 }
 
-int *twoSum(int *nums, int numsSize, int target, int *returnSize);
-
 int main() {
   char line[1024];
   while (fgets(line, sizeof(line), stdin)) {
@@ -366,84 +370,71 @@ int main() {
       free(output);
   }
   return 0;
-}
-
-/**
- * Note: The returned array must be malloced, assume caller calls free().
- */
-int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
-    
 }`,
               },
               {
                 language: "cpp",
-                template: `
-#include <iostream>
-#include <vector>
-#include <string>
+                template: `#include <iostream>
 #include <sstream>
-#include <algorithm>
-#include <unordered_map>
+#include <string>
+#include <vector>
 
 using namespace std;
 
-// 解析输入字符串为整数数组
-vector<int> parseIntArray(string line) {
+class Solution {
+public:
+  vector<int> twoSum(vector<int> &nums, int target) {
+    
+  }
+};
+
+vector<int> parseIntArray(const string &input) {
   vector<int> result;
-    line.erase(remove(line.begin(), line.end(), '['), line.end());
-    line.erase(remove(line.begin(), line.end(), ']'), line.end());
-    stringstream ss(line);
+  string trimmed = input.substr(1, input.size() - 2);
+  stringstream ss(trimmed);
   string token;
+
   while (getline(ss, token, ',')) {
-        if (!token.empty()) {
     result.push_back(stoi(token));
   }
-    }
+
   return result;
 }
 
-// 格式化输出结果为字符串
-string formatOutput(const vector<int>& res) {
-    if (res.empty()) return "[]";
+string formatOutput(const vector<int> &output) {
+  if (output.empty())
+    return "[]";
+
   stringstream ss;
   ss << "[";
-    for (size_t i = 0; i < res.size(); ++i) {
-        ss << res[i];
-        if (i != res.size() - 1)
+
+  for (size_t i = 0; i < output.size(); ++i) {
+    ss << output[i];
+    if (i != output.size() - 1)
       ss << ",";
   }
+
   ss << "]";
   return ss.str();
 }
 
-// Solution 类声明
-class Solution {
-public:
-    vector<int> twoSum(vector<int>& nums, int target);
-};
-
 int main() {
   string line;
+
   while (getline(cin, line)) {
     vector<int> nums = parseIntArray(line);
 
-        if (!getline(cin, line)) break;
+    getline(cin, line);
     int target = stoi(line);
 
     Solution sol;
-        vector<int> res = sol.twoSum(nums, target);
+    vector<int> result = sol.twoSum(nums, target);
 
-        cout << formatOutput(res) << endl;
+    cout << formatOutput(result) << endl;
   }
 
   return 0;
-}
-
-vector<int> Solution::twoSum(vector<int>& nums, int target) {
-
-    return {};
-}
-`,
+}`,
               },
             ],
           },
@@ -625,69 +616,68 @@ $(3 → 4 → 2) + (4 → 6 → 5) = 8 → 0 → 7$`,
             create: [
               {
                 language: "c",
-                template: `
-#include <stdio.h>
+                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;
+struct ListNode *addTwoNumbers(struct ListNode *l1, struct ListNode *l2) {
   
+}
+
+struct ListNode *parseList(char *line) {
   line[strcspn(line, "\\n")] = 0;
   char *p = line;
-    while (*p && (*p == '[' || *p == ' ' || *p == ']')) p++;
+  while (*p && (*p == '[' || *p == ' ' || *p == ']'))
+    p++;
+
+  struct ListNode dummy;
+  struct ListNode *cur = &dummy;
+  dummy.next = NULL;
 
   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;
+    cur->next = node;
+    cur = 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;
+    if (head)
+      printf(",");
   }
   printf("]\\n");
 }
 
-// 释放链表内存
 void freeList(struct ListNode *head) {
   while (head) {
-        struct ListNode* temp = head;
+    struct ListNode *tmp = head;
     head = head->next;
-        free(temp);
+    free(tmp);
   }
 }
 
-// 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 *l1 = parseList(line);
+    if (!fgets(line, sizeof(line), stdin))
+      break;
+    struct ListNode *l2 = parseList(line);
 
     struct ListNode *result = addTwoNumbers(l1, l2);
     printList(result);
@@ -696,29 +686,18 @@ int main() {
     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: `
+                template: `#include <algorithm>
 #include <iostream>
-#include <string>
 #include <sstream>
-#include <vector>
-#include <algorithm>
+#include <string>
+
 using namespace std;
 
-// Definition for singly-linked list.
 struct ListNode {
   int val;
   ListNode *next;
@@ -727,13 +706,13 @@ struct ListNode {
   ListNode(int x, ListNode *next) : val(x), next(next) {}
 };
 
-// 声明 Solution 类
 class Solution {
 public:
-    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2);
+  ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
+    
+  }
 };
 
-// 输入字符串 -> 链表
 ListNode *createList(const string &line) {
   ListNode dummy;
   ListNode *tail = &dummy;
@@ -756,18 +735,17 @@ ListNode* createList(const string& line) {
   return dummy.next;
 }
 
-// 打印链表
 void printList(ListNode *head) {
   cout << "[";
   while (head) {
     cout << head->val;
-        if (head->next) cout << ",";
+    if (head->next)
+      cout << ",";
     head = head->next;
   }
   cout << "]" << endl;
 }
 
-// 释放内存
 void freeList(ListNode *head) {
   while (head) {
     ListNode *tmp = head;
@@ -776,12 +754,12 @@ void freeList(ListNode* head) {
   }
 }
 
-// 主函数
 int main() {
   string line;
   while (getline(cin, line)) {
     ListNode *l1 = createList(line);
-        if (!getline(cin, line)) break;
+    if (!getline(cin, line))
+      break;
     ListNode *l2 = createList(line);
 
     Solution sol;
@@ -793,14 +771,7 @@ int main() {
     freeList(res);
   }
   return 0;
-}
-
-
-ListNode* Solution::addTwoNumbers(ListNode* l1, ListNode* l2) {
-
-    return nullptr; // 在这里填充你的算法逻辑
-}
-`,
+}`,
               },
             ],
           },
@@ -992,91 +963,84 @@ Let $m$ be the size of array \`nums1\` and $n$ be the size of array \`nums2\`.
             create: [
               {
                 language: "c",
-                template: `
-#include <stdio.h>
+                template: `#include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
 
-// 解析输入数组
+double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
+  
+}
+
 int *parseIntArray(char *line, int *len) {
-    line[strcspn(line, "\\n")] = 0; // 移除换行符
+  line[strcspn(line, "\\n")] = 0;
   char *p = line;
-    while (*p && (*p == '[' || *p == ' ' || *p == ']')) p++; // 跳过空格和括号
+  while (*p && (*p == '[' || *p == ' ' || *p == ']'))
+    p++;
 
   int capacity = 10;
-    int *arr = malloc(capacity * sizeof(int)); // 初始分配空间
+  int *arr = malloc(capacity * sizeof(int));
   *len = 0;
 
-    char *token = strtok(p, ","); // 分割输入为逗号分隔的整数
+  char *token = strtok(p, ",");
   while (token) {
-        if (*len >= capacity) { // 扩展数组大小
+    if (*len >= capacity) {
       capacity *= 2;
       arr = realloc(arr, capacity * sizeof(int));
     }
-        arr[(*len)++] = atoi(token); // 存储整数
+    arr[(*len)++] = atoi(token);
     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
+  while (fgets(line, sizeof(line), stdin)) {
+    int nums1Size;
+    int *nums1 = parseIntArray(line, &nums1Size);
 
-        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
+    if (!fgets(line, sizeof(line), stdin))
+      break;
+    int nums2Size;
+    int *nums2 = parseIntArray(line, &nums2Size);
 
-        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
+    double result = findMedianSortedArrays(nums1, nums1Size, nums2, nums2Size);
+    printf("%.5f\\n", result);
 
-        free(nums1); // 释放内存
-        free(nums2); // 释放内存
+    free(nums1);
+    free(nums2);
   }
 
   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>
+                template: `#include <iostream>
 #include <sstream>
-#include <algorithm>
+#include <string>
+#include <vector>
+
 using namespace std;
 
 class Solution {
 public:
-    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2);
+  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> parseVector(const string &line) {
   vector<int> result;
-    stringstream ss(trimmed);
-    string token;
-    while (getline(ss, token, ',')) {
-        if (!token.empty()) {
-            result.push_back(stoi(token));
+  stringstream ss(line);
+  char c;
+  int num;
+  while (ss >> c) {
+    if (isdigit(c) || c == '-' || c == '+') {
+      ss.putback(c);
+      ss >> num;
+      result.push_back(num);
     }
   }
   return result;
@@ -1085,24 +1049,18 @@ vector<int> parseIntArray(const string& line) {
 int main() {
   string line;
   while (getline(cin, line)) {
-        vector<int> nums1 = parseIntArray(line);
-        if (!getline(cin, line)) break;
-        vector<int> nums2 = parseIntArray(line);
+    vector<int> nums1 = parseVector(line);
+
+    if (!getline(cin, line))
+      break;
+    vector<int> nums2 = parseVector(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
-}
-`,
+}`,
               },
             ],
           },