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 };
来源：https://www.cnblogs.com/aezero/p/4823306.html