问题描述：
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: The solution set must not contain duplicate quadruplets.
示例：
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] ]问题分析：
该问题与3Sum问题类似，大家可以先看看我对3Sum的解释。
3Sum问题及解法
详细过程见代码：
class Solution { public: vector<vector<int>> fourSum(vector<int>& nums, int target) { vector<vector<int> > res; if(nums.empty()) return res; sort(nums.begin(), nums.end()); for(int i = 0; i < nums.size(); i++) { int target_3 = target - nums[i]; for(int j = i + 1; j < nums.size(); j++) { int target_2 = target_3 - nums[j]; int front = j + 1; int back = nums.size() - 1; while(front < back) { int two_sum = nums[front] + nums[back]; if(two_sum < target_2) front++; else if(two_sum > target_2) back--; else { vector<int> quadruplet(4, 0); quadruplet[0] = nums[i]; quadruplet[1] = nums[j]; quadruplet[2] = nums[front]; quadruplet[3] = nums[back]; res.push_back(quadruplet); // 保证第三个元素不重复 while(front < back && nums[front] == quadruplet[2]) front++; } } // 保证第二个元素不会相同 while(j + 1 < nums.size() && nums[j + 1] == nums[j]) j++; } // 保证第一个元素不会相同 while(i + 1 < nums.size() && nums[i + 1] == nums[i]) i++; } return res; } };