数组（7）：3Sum

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;
       }
};