Leetcode (15)--3sum

https://leetcode.com/problems/3sum/

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)

Method One:

Two-layer loop + binary lookup, complexity O (n^2 logn). It's too slow.

1 classSolution {2  Public:3vector<vector<int>> Threesum (vector<int>&nums) {4         Set<vector<int>>Res;5vector<vector<int>>output;6vector<int>Sol;7Sort (Nums.begin (), Nums.end ());//Sort First8         inti,j,k,t,x,n=nums.size ();9          for(i=0; i<n;i++){Ten              for(j=i+1; j<n-1; j + +){ Onet=nums[i]+Nums[j]; AX=findsol (-t,nums,j+1, N-1); -                 if(x!=-1){ - sol.clear (); the Sol.push_back (Nums[i]); - Sol.push_back (Nums[j]); - Sol.push_back (nums[x]); - Res.insert (sol); +                 } -             } +         } A         Set<vector<int>>:: iterator iter; at          for(Iter=res.begin (); Iter!=res.end (); iter++){ -Output.push_back (*iter); -         } -         returnoutput; -     } -     intFindsol (inttarget,vector<int> Nums,intBeginintend) { in         if(nums[begin]==target) -             returnbegin; to         if(nums[end]==target) +             returnend;  -         if(nums[begin]>target| | nums[end]<target) the             return-1; *         intmid; $          while(begin<=end) {Panax NotoginsengMid= (begin+end)/2; -             if(nums[mid]==target) the                 returnmid; +             Else if(nums[mid]>target) { Aend=mid-1; the             } +             Else -Begin=mid+1; $         } $         return-1; -     } -};

Method Two: One layer of the loop plus the sum two (https://leetcode.com/problems/two-sum/), O (n^2).

1 classSolution {2  Public:3vector<vector<int>> Threesum (vector<int>&nums) {4 sort (Nums.begin (), Nums.end ());5vector<vector<int>>Res;6         inti,t,a,b,k,n=nums.size ();7          for(i=0; i<n-2; i++){8t=-Nums[i];9a=i+1;Tenb=n-1; One              while(a<b) { Ak=nums[a]+Nums[b]; -                 if(k<t) { -a++; the                 } -                 Else if(k>t) { -b--; -                 } +                 Else{ -vector<int> Sol (3,0); +sol[0]=Nums[i]; Asol[1]=Nums[a]; atsol[2]=Nums[b]; - Res.push_back (sol); -                      while(A < b && nums[a] = = sol[1])  -a++; -                      while(A < b && nums[b] = = sol[2])  -b--; in                  } -             } to              while(i +1< Nums.size () && nums[i +1] ==Nums[i]) +i++; -         } the          *         //Deduplicate, but it's so slow! $         //sort (Res.begin (), Res.end ()); Panax Notoginseng         //res.erase (Unique (Res.begin (), Res.end ()), Res.end ()); -         returnRes; the     } +};

Leetcode (15)--3sum