From d1d3b28341c6f13ca0cc9ee2fd66f5f82af9575c Mon Sep 17 00:00:00 2001
From: Guanzhong Pan <gpan20020205@gmail.com>
Date: Mon, 29 Nov 2021 11:56:50 +0000
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200015.=E4=B8=89=E6=95=B0?=
 =?UTF-8?q?=E4=B9=8B=E5=92=8C.md=20C=E8=AF=AD=E8=A8=80=E7=89=88=E6=9C=AC?=
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---
 problems/0015.三数之和.md | 74 +++++++++++++++++++++++++++++++++++
 1 file changed, 74 insertions(+)

diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md
index dec8a7c6..1e675254 100644
--- a/problems/0015.三数之和.md
+++ b/problems/0015.三数之和.md
@@ -435,5 +435,79 @@ func threeSum(_ nums: [Int]) -> [[Int]] {
 }
 ```
 
+C:
+```C
+//qsort辅助cmp函数
+int cmp(const void* ptr1, const void* ptr2) {
+    return *((int*)ptr1) > *((int*)ptr2);
+}
+
+int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {
+    //开辟ans数组空间
+    int **ans = (int**)malloc(sizeof(int*) * 18000);
+    int ansTop = 0;
+    //若传入nums数组大小小于3，则需要返回数组大小为0
+    if(numsSize < 3) {
+        *returnSize = 0;
+        return ans;
+    }
+    //对nums数组进行排序
+    qsort(nums, numsSize, sizeof(int), cmp);
+    
+
+    int i;
+    //用for循环遍历数组，结束条件为i < numsSize - 2(因为要预留左右指针的位置)
+    for(i = 0; i < numsSize - 2; i++) {
+        //若当前i指向元素>0，则代表left和right以及i的和大于0。直接break
+        if(nums[i] > 0)
+            break;
+        //去重：i > 0 && nums[i] == nums[i-1]
+        if(i > 0 && nums[i] == nums[i-1])
+            continue;
+        //定义左指针和右指针
+        int left = i + 1;
+        int right = numsSize - 1;
+        //当右指针比左指针大时进行循环
+        while(right > left) {
+            //求出三数之和
+            int sum = nums[right] + nums[left] + nums[i];
+            //若和小于0，则左指针+1（因为左指针右边的数比当前所指元素大）
+            if(sum < 0)
+                left++;
+            //若和大于0，则将右指针-1
+            else if(sum > 0)
+                right--;
+            //若和等于0
+            else {
+                //开辟一个大小为3的数组空间，存入nums[i], nums[left]和nums[right]
+                int* arr = (int*)malloc(sizeof(int) * 3);
+                arr[0] = nums[i];
+                arr[1] = nums[left];
+                arr[2] = nums[right];
+                //将开辟数组存入ans中
+                ans[ansTop++] = arr;
+                //去重
+                while(right > left && nums[right] == nums[right - 1])
+                    right--;
+                while(left < right && nums[left] == nums[left + 1])
+                    left++;
+                //更新左右指针
+                left++;
+                right--;
+            }
+        }
+    }
+
+    //设定返回的数组大小
+    *returnSize = ansTop;
+    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
+    int z;
+    for(z = 0; z < ansTop; z++) {
+        (*returnColumnSizes)[z] = 3;
+    }
+    return ans;
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>