4Sum

题目：

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)

思想：

借用3Sum的思想，先固定住前两个数，然后在设定两个指针，根据sum的大小两个指针移动的位置。

代码：

#include <iostream>#include <vector>#include <algorithm>#include <math.h>using namespace std;class Solution {public:    vector<vector<int> > fourSum(vector<int>& nums, int target) {vector<vector<int> >ret;        if(nums.size()<4) return ret;sort(nums.begin(),nums.end());int length=nums.size();int sum;for(int i=0;i<length-3;i++) //确定第一个数 a{if(i>0 &&nums[i]==nums[i-1]){//i++;  //注意此处不要加1，否则会导致错误，满足条件的等式无法输出continue;}int target2=target-nums[i]; //将4Sum转换成3Sum，target2=target-a;for(int j=i+1;j<length-2;++j){if(j>i+1 && nums[j]==nums[j-1])    {   // j++; //同上    continue;    }int k=j+1;int n=length-1;vector<int> tmp;while(k<n){sum=nums[j]+nums[k]+nums[n];if(sum==target2){if(k>j+1 && nums[k]==nums[k-1]){k++;continue;}tmp.clear();tmp.push_back(nums[i]);tmp.push_back(nums[j]);tmp.push_back(nums[k]);tmp.push_back(nums[n]);ret.push_back(tmp);k++;//继续查找}else if(sum>target2){n--;}else{k++;}}}}return ret;    }};int main(){Solution s;int arr[]={0,0,4,-2,-3,-2,-2,-3};vector<int> ivec(arr,arr+8);vector<vector<int> > ret;ret=s.fourSum(ivec,-1);vector<int> tmp;for(auto iter=ret.begin();iter!=ret.end();++iter){tmp.clear();tmp=(*iter);cout<<"("<<tmp[0]<<","<<tmp[1]<<","<<tmp[2]<<","<<tmp[3]<<")"<<endl;}system("pause");return 0;}