LeetCode *** 299. Bulls and Cows

最新推荐文章于 2024-08-05 07:25:56 发布

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最新推荐文章于 2024-08-05 07:25:56 发布

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本文介绍了一个实现Bulls and Cows游戏的算法，包括如何根据秘密数字和玩家猜测提供反馈，以及如何处理重复数字的情况。通过实例演示了算法的应用，并提供了详细代码解析。

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题目：

You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called "bulls") and how many digits match the secret number but locate in the wrong position (called "cows"). Your friend will use successive guesses and hints to eventually derive the secret number.

For example:

Secret number:  "1807"
Friend's guess: "7810"

Hint: 1 bull and 3 cows. (The bull is 8, the cows are 0, 1 and 7.)

Write a function to return a hint according to the secret number and friend's guess, use A to indicate the bulls and B to indicate the cows. In the above example, your function should return "1A3B".

Please note that both secret number and friend's guess may contain duplicate digits, for example:

Secret number:  "1123"
Friend's guess: "0111"

In this case, the 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow, and your function should return "1A1B".

You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal.

分析：

由于secret和guess相同长度，且数字永远只有10个，那么记录secret数字的个数即可。

代码：

class Solution {
public:
    string getHint(string secret, string guess) {
        
        int length=secret.length();
        int flag[10]={};
        
        int A=0,B=0;
        
        for(int i=0;i<length;++i){
            if(secret[i]==guess[i]){
                A++;
                guess[i]=' ';
            }
            else flag[secret[i]-'0']++;
        }
        
        for(int i=0;i<length;++i){
            if(guess[i]!=' '){
                if(flag[guess[i]-'0']){
                    B++;
                    flag[guess[i]-'0']--;
                }
            }
        }
        
        return itos(A)+"A"+itos(B)+"B";
        
    }
    
    string itos(int i){
        string res;
        if(i==0)res="0";
        else {
            while(i){
                res.insert(res.begin(),i%10+'0');
                i/=10;
            }
        }
        return res;
    }
};