15. 3Sum

一开始是用回溯做的，结果超时，代码如下：

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
       vector<vector<int>> ans;
       sort(nums.begin(),nums.end()); //排序，便于剔除重复答案
       vector<int> temp;
       find(nums,ans,temp,0,0);       
       return ans;
        
    }
    void find(vector<int>& nums,vector<vector<int>>& ans, vector<int>& temp,int begin,int sum)
    {
        if(temp.size()==3&&sum==0)
        {         
            ans.push_back(temp);
            return ;
        }
        
        for(int i = begin;i<nums.size();i++)
        {
            if(i>begin&&nums[i]==nums[i-1])
                continue;  //去重
            temp.push_back(nums[i]);
            find(nums,ans,temp,i+1,sum+nums[i]);
            temp.pop_back();
            
        }
    }
};

于是改为时间复杂度为o(n*n)的“双指针”推广法：

class Solution {
public:
   vector<vector<int>> threeSum(vector<int>& nums) {
	 vector<vector<int>> ans;
	 vector<int> temp;
	 int i=0;
       if(nums.size()<3)
           return ans;
	 sort(nums.begin(),nums.end());
	 while(i<nums.size()-2)
	 {
		 int begin=i+1,end=nums.size()-1;
		 while(begin<end)
		 {
			 int sum=nums[i]+nums[begin]+nums[end];
			 if(sum==0)
			 {
				 temp.push_back(nums[i]);
                 temp.push_back(nums[begin]);
                 temp.push_back(nums[end]);
				 ans.push_back(temp);
				 temp.clear();
				 begin++;
				 end--;
				 while(nums[begin]==nums[begin-1])
					 begin++;
				 while(nums[end]==nums[end+1])
					 end--;
			 }
			 else
			 {
				 if(sum<0)
					 begin++;
				 else
					 end--;
			 }				 
		 }		 
		 i++;
		 while(nums[i]==nums[i-1])
			 i++;		 
	 }	 
	 return ans;
 }
};