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)

 C++代码实现：

#include<iostream>

#include<algorithm>

#include<vector>

using namespace std;



class Solution

{

public:

    vector<vector<int> > fourSum(vector<int> &num,int target)

    {

        if(num.empty())

            return vector<vector<int> >();

        sort(num.begin(),num.end());

        vector<vector<int> > ret;

        int n=num.size();

        int i,j;

        for(i=0; i<n-3; i++)

        {

            //只保留第一个不重复的，其余的都删了,因为left会选择重复的

            if(i>=1&&num[i]==num[i-1])

                continue;

            for(j=n-1; j>i+2; j--)

            {

                //只保留最后一个不重复的，其余的都删了,因为right会选择重复的

                if(j<n-1&&num[j+1]==num[j])

                   continue;

                int left=i+1;

                int right=j-1;

                vector<int> tmp;

                while(left<right)

                {

                    //只保留最后一个不重复的，其余的都删了,因为left会选择重复的

                    if(right<j-1&&num[right]==num[right+1])

                        right--;

                    else if(num[i]+num[j]+num[left]+num[right]==target)

                    {

                        tmp= {num[i],num[left],num[right],num[j]};

                        ret.push_back(tmp);

                        left++;

                        right--;

                    }

                    else if(num[i]+num[j]+num[left]+num[right]<target)

                        left++;

                    else if(num[i]+num[j]+num[left]+num[right]>target)

                        right--;

                }

            }

        }

        return ret;

    }

};

int main()

{

    vector<int> vec= {-5,5,4,-3,0,0,4,-2};

    Solution s;

    vector<vector<int> > result=s.fourSum(vec,4);

    for(auto a:result)

    {

        for(auto v:a)

            cout<<v<<" ";

        cout<<endl;

    }

    cout<<endl;

}

注意处理重复的vector。。。。