[leetcode]299. Bulls and Cows

本文介绍了一种实现 Bulls and Cows 游戏提示算法的方法。该算法接收秘密数字和猜测数字作为输入,并返回包含正确位置数字数量(A)和正确数字但错误位置的数量(B)的字符串。通过遍历两个字符串并使用计数数组来跟踪匹配情况,最终计算出 A 和 B 的数量。

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题目链接:https://leetcode.com/problems/bulls-and-cows/#/description

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.

class Solution{
public:
    string getHint(string secret,string guess)
    {
        vector<int> a(10,0);
        vector<int> b(10,0);
        int countA=0,countB=0;
        for(int i=0;i<secret.length();i++)
        {
            if(secret[i]==guess[i])
                countA+=1;
            else
            {
                a[secret[i]-'0']+=1;
                b[guess[i]-'0']+=1;
            }
        }
        for(int i=0;i<10;i++)
        {
            countB+=min(a[i],b[i]);
        }

        string res=to_string(countA)+"A"+to_string(countB)+"B";
        return res;
    }
};




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