1、twoSum

关键：Hash

unordered_map<int, int> map;

map.count(key) 判断key对应value的个数

map[key] = value 插入{key,value}

复杂度为O(n)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> hash;
        vector<int> res;
     for(int i=0; i<nums.size(); i++){
         int y = target - nums[i];
         if(hash.count(y)){
             res.push_back(i);
             res.push_back(hash[y]);
         }
         hash[nums[i]]=i;
     }  
     return res;
    }
};