LeetCode - 18. 4Sum

18. 4Sum

Problem's Link

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[b]Mean:[/b]

给定一个数列和一个目标数，找出所有和为目标数的四元组.

[b]analyse:[/b]

和3Sum差不多，多加了一个循环而已.

[b]Time complexity: O(N^3)[/b]

[b]view code[/b]

/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-02-17-13.10
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);

class Solution
{
public:
vector<vector<int>> fourSum(vector<int>& nums, int target)
{
vector<vector<int>> res;
int si=nums.size();
if(si<4) return res;
std::sort(nums.begin(),nums.end());
int target_3,target_2;
int low,high;
for(int i=0;i<si;++i)
{
target_3=target-nums[i];
for(int j=i+1;j<si;++j)
{
target_2=target_3-nums[j];
low=j+1;
high=si-1;
while(low<high)
{
int two_sum=nums[low]+nums[high];
if(two_sum<target_2)
++low;
else if(two_sum>target_2)
--high;
else
{
vector<int> quadruplet(4,0);
quadruplet[0]=nums[i];
quadruplet[1]=nums[j];
quadruplet[2]=nums[low];
quadruplet[3]=nums[high];
res.push_back(quadruplet);
while(low<high && nums[low]==quadruplet[2]) ++low;
while(low<high && nums[high]==quadruplet[3]) --high;
}
}
while(j+1<si && nums[j+1]==nums[j]) ++j;
}
while(i+1<si && nums[i+1]==nums[i]) ++i;
}
return res;
}
};

int main()
{
Solution solution;
int n,target;
while(cin>>n>>target)
{
vector<int> ve;
for(int i=0;i<n;++i)
{
int tmp;
cin>>tmp;
ve.push_back(tmp);
}
auto ans=solution.fourSum(ve,target);
for(auto p1:ans)
{
for(auto p2:p1)
cout<<p2<<" ";
cout<<endl;
}
}
return 0;
}
/*

*/