Leetcode题解 299. Bulls and Cows

本文介绍了一个用于Bulls和Cows猜数字游戏的提示生成算法。该算法通过比较秘密数字与猜测数字来计算正确位置上的数字数量(称为‘Bulls’)以及正确数字但位置错误的数量(称为‘Cows’),并返回相应的提示。

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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.

逐位比较,相等的直接+1,不等的分别存起来,最后单独比较不等的这部分,看看有哪些数字是错位的。

public class Solution {
    public String getHint(String secret, String guess) {
        int Anum=0;
        int Bnum=0;
        int[] resultA=new int[10];
        int[] resultB=new int[10];
        for(int i=0;i<secret.length();i++){
            int tempA=secret.charAt(i)-48;
            int tempB=guess.charAt(i)-48;
            if(tempA==tempB){
                Anum++;
            }else{
                resultA[secret.charAt(i)-48]++;
                resultB[guess.charAt(i)-48]++;
            }
        }
        for(int i=0;i<10;i++){
            if(resultA[i]!=0&&resultB[i]!=0){
                Bnum+=resultA[i]>resultB[i]?resultB[i]:resultA[i];
            }
        }
        return Anum+"A"+Bnum+"B";
    }
}
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