Leetcode_c++:3sum (015)

• Topic
  Given an array S of n integers, is 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 is 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)

• Complexity: O (n*n).

/ * First sort, then left and right clamp O (n*n) =sort (NLGN) +n*twopointers (n): O (n*n) * /#include <iostream>#include <vector>#include <algorithm>using namespace STD;classsolution{ Public: vector<vector<int> >Threesum ( vector<int>&num) { vector<vector<int> >RetintLen=num.size ();intTar=0;if(len<=2)returnRet Sort (Num.begin (), Num.end ()); for(intI=0; i<=len-3; i++) {//first Number:num[1]            intj=i+1;//second Number            intk=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; }        }returnRet }};intMain () { vector<int>NumintN,t; while(Cin>>n) { for(intI=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; }}

Leetcode_c++:3sum (015)