9 3Sum

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:[ [-1, 0, 1] , [-1, -1, 2] ]

一开始想用暴力法先提交个答案，没想到超时了。后来参照twoSum又写了一版，用map查找第三个数，没想到又超时了。。。贴一下我的超时代码：

class Solution {
public:
     vector<vector<int>> threeSum(vector<int>& nums) {
        
        vector<vector<int>> ans;
        
        if(nums.size()<3) return ans;
        
        map<int,int> mydict;
        
        for(int i=0;i<nums.size();i++)
        {
            mydict[nums[i]] = i;
        }
        
        
        //brute force
        for(int i=0;i<nums.size();i++)
        {
            for(int j=i+1;j<nums.size();j++)
            {
                int res =  0 - nums[i] - nums[j];
                
                auto it = mydict.find(res);
                if((it != mydict.end()) && (it->second > j))
                {
                    vector<int> temp;
                        
                    temp.push_back(nums[i]);
                    temp.push_back(nums[j]);
                    temp.push_back(res);
                    
                    std::sort(temp.begin(),temp.end());
                        
                    ans.push_back(temp);  
                }
                
            }
        }
        
        //delete duplicates
        std::sort(ans.begin(),ans.end());
        ans.erase(unique(ans.begin(),ans.end()), ans.end());
        
        
        return ans;  
    }
};

后来想了很久也没什么特别的思路，决定去讨论区参照一下已有的算法，附上参考的算法链接：https://leetcode.com/problems/3sum/discuss/274024/C%2B%2B-93.31100ms100(14.5M)-.
这个算法的新奇之处在于在将原有数组进行了从小到大的排序，这样通过从排序后的数组两头逼近来求取和为零的组合，真是神操作（速度超越95%）。

class Solution {
public:
     vector<vector<int>> threeSum(vector<int>& nums) {
        
        vector<vector<int>> ans;
        
        if(nums.size()<3) return ans;
        
        //sort
        std::sort(nums.begin(),nums.end());
        int len = nums.size();
        
        //
        for(int i=0;i<len-2;i++)
        {
          
            if(i>0 && nums[i] == nums[i-1]) continue;
            
            if(nums[i] + nums[i+1] + nums[i+2] > 0) break;
            if(nums[i] + nums[len-2] + nums[len-1] <0) continue;
            
            int j=i+1, k = len -1;
            while(j<k)
            {
                int sum = nums[i] + nums[j] + nums[k];
                if(sum > 0) k--;
                else if(sum<0) j++;
                else{
                    ans.push_back({nums[i],nums[j],nums[k]});
                    do{j++;}while(nums[j] == nums[j-1] && j<k );
                    do{k--;}while(nums[k] == nums[k+1] && j<k );
                }
            }    
            
        }
        
        
        return ans;  
    }
};