LeetCode/solutions/299. Bulls and Cows.md

# [299. Bulls and Cows](https://leetcode.com/problems/bulls-and-cows/)

# 思路
A的个数比较简单，只需要挨个位置比较即可。B的个数，需要先用一个数组记录secret中各数字出现的个数（需排除A情况），再遍历一遍看guess中的数字是否在secret中出现过，若是，则B个数加一。


# C++


```C++
class Solution {
public:
    string getHint(string secret, string guess) {
        int a = 0, b = 0;
        vector<int>mp(10, 0);
        for(int i = 0; i < secret.size(); i++){
            if(secret[i] != guess[i]) mp[secret[i] - '0']++;
        }
        for(int i = 0; i < secret.size(); i++){
            if(secret[i] == guess[i]) a++;
            else if(mp[guess[i] - '0'] > 0){
                b++;
                mp[guess[i] - '0']--;
            }
        }
        
        return to_string(a) + "A" + to_string(b) + "B";
    }
};
```
也可以只遍历一遍：
```C++
class Solution {
public:
    string getHint(string secret, string guess) {
        int a = 0, b = 0;
        vector<int>mp1(10, 0);
        vector<int>mp2(10, 0);
        for(int i = 0; i < secret.size(); i++){
            if(secret[i] != guess[i]){
                mp1[secret[i] - '0']++;
                mp2[guess[i] - '0']++;
            }else{
                a++;
            }

        }
        for(int i = 0; i < 10; i++){
            b += min(mp1[i], mp2[i]);
        }
        
        return to_string(a) + "A" + to_string(b) + "B";
    }
};
```