描述：
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)分析：
先排序，后左右夹逼，复杂度O(n^2)。
这个方法可以推广到K-sum,先排序，然后做k-2次循环，在最内层循环左右夹逼，时间复杂度 O(max{nlogn,n^k?1 })。
代码
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());const int target = 0;auto last = nums.end();for (auto i = nums.begin(); i < last-2; ++i) {auto j = i+1;if (i > nums.begin() && *i == *(i-1)) continue;auto k = last-1;while (j < k) {if (*i + *j + *k < target) {++j;while(*j == *(j - 1) && j < k) ++j;} else if (*i + *j + *k > target) {--k;while(*k == *(k + 1) && j < k) --k;} else {result.push_back({ *i, *j, *k });++j;--k;while(*j == *(j - 1) && *k == *(k + 1) && j < k) ++j;}}}return result;}
};