1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

Solution

1. brute force

class Solution {   public:    vector
   
     twoSum(vector
    
     & nums, int target) {           int n = nums.size();        vector
     
       res;         for (int i = 0; i < n; i++)        {               for (int j = i+1; j < n; j++)            {                   if(nums[i]+nums[j] == target) {                       res.push_back(i);                    res.push_back(j);                    return res;                }             }        }        return res;    }};

2. hashmap

下面要用到unordered_map

unordered_map
   
    ::iterator it;it->first;               // same as (*it).first   (the key value)it->second;              // same as (*it).second  (the mapped value)

代码：

class Solution {   public:    vector
   
     twoSum(vector
    
     & nums, int target) {   	unordered_map
     
       hmap;	vector
      
        res;	for ( int i = 0; i < nums.size(); i++)	{   		int input = nums[i]; //避免多次查找元素，先储存下来		int key = target - input; //计算关键码		auto it = hmap.find(key); // auto 简化变量初始化，根据后面的数据类型推测；find 通过给定关键码查找元素，如果key存在，则find返回key对应的迭代器，如果key不存在，则find返回unordered_map::end		if ( it != hmap.end()) //如果key存在，说明已经找到符合条件的值，终止循环并返回		{   			res.push_back(it->second); 	//保存哈希表中的mapped value			res.push_back(i);	//保存当前下标			break;		}		//如果没有找到key，继续建立hash table		hmap[input] = i; //保存nums[i]关键码所对应的下标值	}	return res;}};