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4Sum

2024-08-04 07:39:57   217人阅读

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。。。。

 

4Sum


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