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3Sum

2024-08-04 06:33:55   218人阅读

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

  • Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
  • The solution set must not contain duplicate triplets.

 

    For example, given array S = {-1 0 1 2 -1 -4},    A solution set is:    (-1, 0, 1)    (-1, -1, 2)

这题做得真是郁闷。。。。

C++实现代码：

#include<iostream>#include<algorithm>#include<vector>using namespace std;class Solution{public:    vector<vector<int> > threeSum(vector<int> &num)    {        if(num.empty())            return vector<vector<int> >();        sort(num.begin(),num.end());        vector<vector<int> > ret;        int n=num.size();        int i;        for(i=0; i<n-2; i++)        {            //只保留第一个不重复的，其余的都删了            if(i>=1&&num[i]==num[i-1])                continue;            int target=-num[i];            int left=i+1;            int right=n-1;            vector<int> tmp;            while(left<right)            {                //删除重复元素                if(right>left&&right<n-1&&num[right+1]==num[right])                    right--;                else if(num[left]+num[right]==target)                {                    tmp={num[i],num[left],num[right]};                    ret.push_back(tmp);                    left++;                    right--;                }                else if(num[left]+num[right]<target)                    left++;                else if(num[left]+num[right]>target)                    right--;            }        }        return ret;    }};int main(){    vector<int> vec= {-2,0,1,1,1,1,1,2};    Solution s;    vector<vector<int> > result=s.threeSum(vec);    for(auto a:result)    {        for(auto v:a)            cout<<v<<" ";        cout<<endl;    }    cout<<endl;}

 

 

3Sum


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