18.4Sum

思路：

• 类似3Sum，利用map存储可以省去判断重复数组，复杂度为 (O(n^4))，因为set判断元素是否重复也需要 (O(n)).

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
    	sort(nums.begin(), nums.end());
       vector<vector<int>> res;
       set<vector<int>> tmp;
       int len = nums.size();
       for(int i = 0; i < len - 3; i++) {
       		for(int j = i+1; j < len -2 ; j++){
       			int lo = j+1,hi = len-1;
       			while(lo < hi){
       				if(nums[i] + nums[j] + nums[hi] + nums[lo] == target){
       					tmp.insert({nums[i],nums[j],nums[lo],nums[hi]});
       					lo++;
       					hi--;
       				}else if(nums[i] + nums[j] + nums[hi] + nums[lo] < target){
       					lo++;
       				}else hi--;
       			}
       		}
       }
       for(auto m : tmp){
       		res.push_back(m);
       }
       return res;
    }
};