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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

   先用死办法试试看，o(n^2) ，很慢！

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> p(2,0);
        for(int i = 0; i < nums.size(); ++i)
        {
            for(int j = i+1; j < nums.size(); ++j)
            {
                if(nums[i] + nums[j] == target)
                {
                    p[0] = i;
                    p[1] = j;
                    break;
                }
                
                
            }
            
        }
        return p;
        
    }
};

    hash map 的 用法，神奇，很快！

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target)
{
    //Key is the number and value is its index in the vector.
	unordered_map<int, int> hash;
	vector<int> result;
	for (int i = 0; i < numbers.size(); i++) {
		int numberToFind = target - numbers[i];

            //if numberToFind is found in map, return them
		if (hash.find(numberToFind) != hash.end()) {
            
			result.push_back(hash[numberToFind]);
			result.push_back(i);			
			return result;
		}

            //number was not found. Put it in the map.
		hash[numbers[i]] = i;
	}
	return result;
}
};

    也是map，  稍微更好理解一点的~

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
       unordered_map<int,int> mp;
       vector<int> result;
       
       for(int i=0;i<nums.size();i++){
           mp[nums[i]]=i;
       }
       
       for(int i=0;i<nums.size();i++){
           if(mp.find(target-nums[i])!=mp.end() && mp[target-nums[i]]!=i) {
               result.push_back(i);
               result.push_back(mp[target-nums[i]]);
               break;
           }
       }
       
       return result;
    }
};

      second 的 用法。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> map;
        vector<int> p(2,0);
        int n = (int)nums.size();
        for (int i = 0; i < n; i++) {
            auto p = map.find(target-nums[i]);
            if (p!=map.end()) {
                return {p->second, i};  //大括号！
            }
            map[nums[i]]=i;
        }
        return p;
    }
};