E299：Bulls and Cows（字符串处理）

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本文介绍了一个用于实现经典猜数字游戏Bulls and Cows的提示生成算法。该算法能够准确地计算出玩家猜测的数字与秘密数字之间的匹配情况，包括位置正确的数字数量（Bulls）及位置错误但数字正确的数量（Cows）。通过逐个字符的比对，并使用HashMap来跟踪每个字符出现的次数，确保了算法的有效性和准确性。

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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.

大意就是给两个字符串，输出两个参数：

① 在两个字符串中位置相同且数值相等的数字的个数

② 除了①的匹配外，两个字符串数值相等但是位置不相同的其他数字个数

public String getHint(String secret, String guess) {
        	
        	HashMap s=new HashMap();//用两个HashMap分别存储两个字符串的字符
        	HashMap g=new HashMap();
        	int a=0;//a记录①条件的数值
        	int b=0;//b记录②条件的数值
        	for(int i=0;i<secret.length();i++){
        		s.put(i+1, secret.charAt(i));
        	}
        	for(int i=0;i<guess.length();i++){
        		g.put(i+1, guess.charAt(i));
        	}
        	for(int i=1;i<=secret.length();i++){
        		if(s.get(i).equals(g.get(i))){
        			a++;
        		}
        	}
        	for(int i=1;i<=secret.length();i++){
        		for(int j=1;j<=secret.length();j++)
        			if(s.get(i).equals(g.get(j))){
        				g.put(j, null);//这里比较重要，对比完一次就删掉一个字符，防止重复
        				b++;
        				break;
        			}
        	}
            return a+"A"+(b-a)+"B";
    }