问题描述：点击打开链接 leetcode
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)
解析：
首先排序，然后左右夹逼，复杂度O(n^2) -- 从程序中可以看出
此方法可扩展到 k-sum （即求k个数的和等于target）,先排序，然后做 k-2 次循环，在最内层循环中左右夹逼；时间复杂度为<math>max(O(nlogn),O(n^{k - 1}))</math>
（排序<math>O(nlogn)</math>, 循环<math>O(n^{k - 1})</math>
#include <iostream>
#include <vector>
#include <algorithm>using namespace std;
class Solution {
public:vector<vector<int>> threeSum(vector<int>& nums) {vector< vector<int> > result;if (nums.size() < 3)return result;sort(nums.begin(), nums.end());int target = 0;for (int i = 0; i < nums.size()-2; i++) {int j = i+1;int k = nums.size() - 1;if (i > 0 && nums[i] == nums[i-1])continue;while (j < k) {int sum = nums[i] + nums[j] + nums [k];if (sum < target) {j++;while (j < k && nums[j] == nums[j-1])j++;} else if (sum > target) {k--;while (j < k && nums[k] == nums[k+1])k--;} else {int t[] = {nums[i], nums[j], nums[k]};vector<int> temp(t, t+3);result.push_back(temp);cout << nums[i] << " "<< nums[j] << " " << nums[k] << endl;j++;k--;while (j < k && nums[j] == nums[j-1] && nums[k] == nums[k+1])//只有两个都等时，需要改变一个j++;}}}return result;}};
int main()
{int test1[] = {-1, 0, 1, 2, -1, -4};int test2[] = {-1, 0, 1};int test4[] = {-1, 0};int test3[] = {2, 3, -2, 0, -10, 4, 6};vector<int> nums(test1, test1+6);vector<int> nums2(test2, test2+3);vector<int> nums3(test3, test3+7);vector<int> nums4(test4, test4+2);Solution sol;sol.threeSum(nums);sol.threeSum(nums2);sol.threeSum(nums3);sol.threeSum(nums4);
}