4SUM

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)

解题思路：一般解决2SUM问题时，采用哈希来做，3sum问题则是固定两个数值，然后遍历第三个来寻找；而这个问题也是同样的思路，首先对数组排序，固定两个数值，通过比较4个数和的大小与目标值的大小关系调整三四个数的索引，注意这里不允许重复答案，所以建一个映射表来表示已找到的答案值；

#include<iostream>#include<vector>#include<algorithm>#include<set>using namespace std;vector<vector<int> >fourSum(vector<int> &num,int target ){vector<vector<int> >ResultVector;set<vector<int> >   ResultSet;if (num.size() < 4)return ResultVector;sort(num.begin(), num.end());for (int idx_fis = 0; idx_fis != num.size();++idx_fis){for (int idx_sec = idx_fis + 1; idx_sec != num.size();++idx_sec){int idx_thir = idx_sec + 1;int idx_four = num.size() - 1;while (idx_thir<idx_four){int sum = num[idx_fis] + num[idx_sec] + num[idx_thir] + num[idx_four];if (sum == target){vector<int>TmpResult;TmpResult.push_back(num[idx_fis]);TmpResult.push_back(num[idx_sec]);TmpResult.push_back(num[idx_thir]);TmpResult.push_back(num[idx_four]);if (!ResultSet.count(TmpResult)){ResultVector.push_back(TmpResult);ResultSet.insert(TmpResult);}idx_thir++;idx_four--;}else if (sum < target)idx_thir++;elseidx_four--;}}}return ResultVector;}