1. 题目描述

给你一个包含 n 个整数的数组 nums，判断 nums 中是否存在三个元素 a，b，c ，使得 a + b + c = 0 ？请你找出所有满足条件且不重复的三元组。注意：答案中不可以包含重复的三元组。示例：给定数组 nums = [-1, 0, 1, 2, -1, -4]，满足要求的三元组集合为：[ [-1, 0, 1], [-1, -1, 2]]2. 测试结果

3. 解法1_双指针法

/*author: xidoublestardate: 2020-5-21*/int compare(const void* a, const void* b){ return (*(int*)a - *(int*)b);}int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) { //排除平台只输出]的bug *returnSize = 0; //排除输入小于3的情况 if (numsSize < 3) return NULL; //使用qsort函数快速排序 qsort(nums, numsSize, sizeof(int), compare); //申请二级指针空间 int** returnArray = (int**)malloc(sizeof(int*) * (numsSize - 2) * (numsSize - 2)); //申请每个一维数组大小的空间 *returnColumnSizes = (int*)malloc(sizeof(int) * (numsSize - 2) * (numsSize - 2)); int cur = 0, low = 0,high = 0; //cur遍历数组，low和high分别做为左值和右值的下标往中间夹 while (nums[cur] <= 0 && cur < numsSize - 2) { low = cur + 1; high = numsSize - 1; while (low < high) { if (0 == (nums[cur] + nums[low] + nums[high])) { returnArray[*returnSize] = (int*)malloc(sizeof(int) * 3);//每找到一组，二级指针分配3个空间 (*returnColumnSizes)[*returnSize] = 3;//记录列数 returnArray[*returnSize][0] = nums[cur]; returnArray[*returnSize][1] = nums[low]; returnArray[(*returnSize)++][2] = nums[high]; while ((nums[low] == nums[++low]) && (low < high));//往后去重 while ((nums[high] == nums[--high]) && (low < high));//往前去重 } else if (0 < (nums[cur] + nums[low] + nums[high])) { high--; } else { low++; } } while ((nums[cur] == nums[++cur]) && (cur < numsSize - 2));//cur左值去重 } return returnArray;}4、复杂度分析

时间复杂度：O(n*n)
空间复杂度：O(1)