leetcode_c++：3Sum(015)

• 题目
  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)

• 复杂度： O(n*n)。

/* 先排序，然后左右夹逼 O(n*n)=sort(nlgn)+n*twopointers(n):O(n*n) */ #include<iostream> #include<vector> #include <algorithm> using namespace std; class Solution{ public: vector<vector<int> > threeSum(vector<int> &num){ vector<vector<int> > ret; int len=num.size(); int tar=0; if(len<=2) return ret; sort(num.begin(),num.end()); for(int i=0;i<=len-3;i++){ //first number:num[1] int j=i+1; //second number int k=len-1; //third number while(j<k){ if(num[i]+num[j]+num[k]<tar){ ++j; }else if(num[i]+num[j]+num[k]>tar){ --k; }else{ vector<int> temp; temp.push_back(num[i]); temp.push_back(num[j]); temp.push_back(num[k]); ret.push_back(temp); ++j; --k; //follow 3 while can avoid the duplication while(j<k && num[j]==num[j-1]) ++j; while(j<k && num[i]==num[i+1]) ++i; } } while ((i<len-3) && num[i] == num[i+1]) { ++i; } } return ret; } }; int main() { vector<int> num; int n,t; while(cin>>n){ for(int i=0;i<n;i++){ cin>>t; num.push_back(t ); } Solution s; vector<vector<int> > ret=s.threeSum(num); for(vector<vector<int> >::iterator it=ret.begin();it!=ret.end();it++){ for(vector<int>::iterator it1=it->begin();it1!=it->end();it1++) cout<<*it1<<" "; cout<<endl; } return 0; } }