【LeetCode】4Sum

题目描述：

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

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.    A solution set is:    (-1,  0, 0, 1)    (-2, -1, 1, 2)    (-2,  0, 0, 2)

一开始想用两层墙壁往中间挤的方式来做，这样只有O（n^2）的复杂度，但是后来发现这样会漏掉一些组合。

最终只能确定最外面两层循环，最里一层往中间挤。跟3Sum Closest类似，不过多了一层循环。

代码如下：

class Solution {public:vector<vector<int> > fourSum(vector<int> &num, int target) {if (num.size() < 4)return vector<vector<int>>();sort(num.begin(), num.end());vector<vector<int>> res;int N = num.size();for (int i = 0; i < N; i++){if (i - 1 >= 0 && num[i] == num[i - 1])continue;for (int j = i + 1; j < N; j++){if (j != i + 1 && num[j] == num[j - 1])continue;int l(j + 1), r(N - 1);while (l < r){if (l != j + 1 && num[l] == num[l - 1]){l++;continue;}if (r + 1 < N&&num[r] == num[r + 1]){r--;continue;}int sum = num[i] + num[j] + num[l] + num[r];if (sum == target)res.push_back(vector<int>({ num[i], num[j], num[l], num[r] }));if (sum < target)l++;elser--;}}}return res;}};