015 - 3Sum

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:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie,a ≤ b ≤ c)
• 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)

给定一个数组，找出这些元素，三个相加等于0的左右组合

int mycompar(const void *a, const void *b){int *x = (int *)a;int *y = (int *)b;if (*x > *y) return 1;return *x == *y? 0 : -1;}int search(int *num, int size, int aim, int i1, int i2){int left = 0, right = size - 1, mid;while (left <= right) {mid = (left + right) / 2;if (num[mid] == aim) {int x = mid, y = mid, i;while (x >= 0 && num[x] == aim) x--;while (y < size && num[y] == aim) y++;for (i = x + 1; i < y; i++)if (i != i1 && i != i2)return i;break;}if (num[mid] > aim) right = mid - 1;else left = mid + 1;}return -1;}int** threeSum(int* nums, int numsSize, int* returnSize) {int i, j, k = 0;int **out = (int **)calloc(sizeof(int *), 1024);for (i = 0; i < 1024; i++)out[i] = (int *)calloc(sizeof(int), 3);qsort(nums, numsSize, sizeof(int), mycompar);for (i = 0; nums[i] <= 0; i ++) {while (i && i < numsSize - 2 && nums[i - 1] == nums[i]) i++;if (i == numsSize - 2) break;for (j = i + 1; j < numsSize - 1; j ++) {while (j-1 > i && j < numsSize - 1 && nums[j] == nums[j - 1]) j++;if (j == numsSize - 1) break;if (-nums[i] - nums[j] < nums[j]) break;if (search(nums, numsSize, -nums[i] - nums[j], i, j) >= 0) {out[k][0] = nums[i];out[k][1] = nums[j];out[k++][2] = -nums[i] - nums[j];}}}*returnSize = k;return out;}