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Given an array S of n integers, are there elements a,b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],A solution set is:[  [-1, 0, 1],  [-1, -1, 2]]

======  参考的网上的思想， 自己写了一遍，Leecode还是报 runtime error ，不知道哪里错了， 接着再继续更改 ====

/** * Return an array of arrays of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */int** threeSum(int* nums, int numsSize, int* returnSize) {        int** ret;        int cnt = 0, left=0, right=0, i , j, tmp=0, sum=0;        if (numsSize < 3) return NULL;                if (numsSize == 3 ) {            if (nums[0]+nums[1]+nums[2] == 0) {                        printf("hello\n");                                ret = (int(*)[3])malloc(sizeof(int*) * 1);                *ret=(int *)malloc(sizeof(int)* 3);                ret[0][0]=nums[0];                ret[0][1]=nums[1];                ret[0][2]=nums[2];                printf("%d, %d, %d\n", ret[0][0], ret[0][1], ret[0][2]);                                *returnSize=numsSize;                printf("%d", *returnSize);                                return ret;                             } else {                return NULL;            }                        }                              int *num1=malloc(sizeof(int) * numsSize *10);        int *num2=malloc(sizeof(int) * numsSize *10);        int *num3=malloc(sizeof(int) * numsSize *10);        memset(num1, 0 , numsSize);        memset(num2, 0 , numsSize);        memset(num3, 0 , numsSize);                for (i=0;i
   
    nums[j]) {                    tmp=nums[i];                    nums[i]=nums[j];                    nums[j]=tmp;                }            }        }            for(i = 0; i < numsSize - 2; i++){            left = i + 1;            right = numsSize - 1;                        while(left < right){                 sum = nums[i] + nums[left] + nums[right];                // 如果三个数的和大于等于目标数，那将尾指针向左移                if(sum > 0) {                    right--;                // 如果三个数的和小于目标数，那将头指针向右移                } else {                    if (sum < 0) {                        left++;                    } else {                        num1[cnt]=nums[i];                        num2[cnt]=nums[left];                        num3[cnt]=nums[right];                                                printf("%d,%d,%d====\n", num1[cnt], num2[cnt], num3[cnt]);                        cnt++;                        left++;                        right--;                        break;                    }                                    }            }        }        if (cnt==0) return NULL;        ret = (int** )malloc( numsSize *2 * sizeof(int*));        for (i=0; i < cnt; i++){            ret[i]=malloc(sizeof(int) * 3);        }        printf("%d\n", cnt);                for (i=0; i
   

====================  更改后， leetcode accept ======================

/** * Return an array of arrays of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */int** threeSum(int* nums, int numsSize, int* returnSize) {        int** ret;        int cnt = 0, left=0, right=0, i , j, tmp=0, sum=0, is_exist=0, p=0, q=0;               if (numsSize < 3) return NULL;              int *num1=malloc(sizeof(int) * numsSize *20);        int *num2=malloc(sizeof(int) * numsSize *20);        int *num3=malloc(sizeof(int) * numsSize *20);        memset(num1, 0 , numsSize);        memset(num2, 0 , numsSize);        memset(num3, 0 , numsSize);                for (i=0;i
   
    nums[j]) {                    tmp=nums[i];                    nums[i]=nums[j];                    nums[j]=tmp;                }            }        }            for(i = 0; i < numsSize - 2; i++){            left = i + 1;            right = numsSize - 1;                        while(left < right){                 sum = nums[i] + nums[left] + nums[right];                // 如果三个数的和大于等于目标数，那将尾指针向左移                if(sum > 0) {                    right--;                // 如果三个数的和小于目标数，那将头指针向右移                } else {                    if (sum < 0) {                        left++;                    } else {                        is_exist=0;                        for (p=0; p
   

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