https://leetcode.com/problems/3sum/
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)
方法一：
两层循环+二分查找，复杂度O(n^2 logn). 太慢了
1 class Solution { 2 public: 3 vector<vector<int>> threeSum(vector<int>& nums) { 4 set<vector<int>> res; 5 vector<vector<int>> output; 6 vector<int> sol; 7 sort(nums.begin(),nums.end()); //sort first 8 int i,j,k,t,x,n=nums.size(); 9 for(i=0;i<n;i++){ 10 for(j=i+1;j<n-1;j++){ 11 t=nums[i]+nums[j]; 12 x=findSol(-t,nums,j+1,n-1); 13 if(x!=-1){ 14 sol.clear(); 15 sol.push_back(nums[i]); 16 sol.push_back(nums[j]); 17 sol.push_back(nums[x]); 18 res.insert(sol); 19 } 20 } 21 } 22 set<vector<int>> :: iterator iter; 23 for(iter=res.begin();iter!=res.end();iter++){ 24 output.push_back(*iter); 25 } 26 return output; 27 } 28 int findSol(int target,vector<int> nums,int begin,int end){ 29 if(nums[begin]==target) 30 return begin; 31 if(nums[end]==target) 32 return end; 33 if(nums[begin]>target||nums[end]<target) 34 return -1; 35 int mid; 36 while(begin<=end){ 37 mid=(begin+end)/2; 38 if(nums[mid]==target) 39 return mid; 40 else if(nums[mid]>target){ 41 end=mid-1; 42 } 43 else 44 begin=mid+1; 45 } 46 return -1; 47 } 48 };
方法二： 一层循环加two sum思想（https://leetcode.com/problems/two-sum/）,O(n^2).
1 class Solution { 2 public: 3 vector<vector<int>> threeSum(vector<int>& nums) { 4 sort(nums.begin(),nums.end()); 5 vector<vector<int>> res; 6 int i,t,a,b,k,n=nums.size(); 7 for(i=0;i<n-2;i++){ 8 t=-nums[i]; 9 a=i+1; 10 b=n-1; 11 while(a<b){ 12 k=nums[a]+nums[b]; 13 if(k<t){ 14 a++; 15 } 16 else if(k>t){ 17 b--; 18 } 19 else{ 20 vector<int> sol(3,0); 21 sol[0]=nums[i]; 22 sol[1]=nums[a]; 23 sol[2]=nums[b]; 24 res.push_back(sol); 25 while (a < b && nums[a] == sol[1]) 26 a++; 27 while (a < b && nums[b] == sol[2]) 28 b--; 29 } 30 } 31 while (i + 1 < nums.size() && nums[i + 1] == nums[i]) 32 i++; 33 } 34 35 //deduplicate, but it is so slow! 36 //sort(res.begin(), res.end()); 37 //res.erase(unique(res.begin(), res.end()), res.end()); 38 return res; 39 } 40 };