翻译给定一个有n个数字的数组S，在S中是否存在元素a，b，c和d的和恰好满足a + b + c + d = target。找出数组中所有的不想等的这四个元素，其和等于target。备注：在（a，b，c，d）中的元素必须从小到大排列。（a ≤ b ≤ c ≤ d）其结果必须不能够重复。例如，给定S = {-2 2}，，target = 0。一个结果集为：(-1, 0, 0, 1)(-2, -1, 1, 2)(-2, 0, 0, 2)原文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 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 = {-2 2}, and target = 0.A solution set is:(-1, 0, 0, 1)(-2, -1, 1, 2)(-2, 0, 0, 2)代码
具体的方法和前面两道题一样，我就不再赘述了。
class Solution {public:vector<vector<int>> fourSum(vector<int>& nums,int target) {sort(nums.begin(), nums.end());vector<vector<int>> result;int len = nums.size();for (int current = 0; current < len – 3;current++){for(int second = current+1;second<len-2;second++){int front = second + 1, back = len – 1;while (front < back){if (nums[current]+nums[second] + nums[front] + nums[back] < target)front++;else if (nums[current] +nums[second]+ nums[front] + nums[back] > target)back–;else{vector<int> v(4);v[0]=nums[current];v[1]=nums[second];v[2]=nums[front];v[3]=nums[back];result.push_back(v);do {front++;} while (front < back&&nums[front – 1] == nums[front]);do {back–;} while (front < back&&nums[back + 1] == nums[back]);}}while(second < len-2&&nums[second+1]==nums[second])second++;}while (current < len – 3 && nums[current + 1] == nums[current])current++;}return result;}};
和本道题关联密切的题目推荐：
传送门：LeetCode 15 3Sum（3个数的和） 传送门：LeetCode 16 3Sum Closest（最接近的3个数的和）
成功不是将来才有的，而是从决定去做的那一刻起，持续累积而成。
