Java [Leetcode 229]Bulls and Cows

Posted 2020-06-13

题目描述：

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.

 

解题思路：

对于secret列每次读取相应的部分，如果与guess对应的部分相同，则bulls变量加一；否则，对于secret的部分相应字典的数值加一，guess对应的字典减一。

再判断secret对应字典的数值是否小于等于0，若是则把bulls的变量加一（表明之前guess中已经出现该字符，且配对成功）

判断guess对应字典的数值是否大于等于0，若是则把bulls的变量加一（表明之前secret中已经出现该字符，且配对成功）

代码如下：

public class Solution {
    public String getHint(String secret, String guess) {
        int bulls = 0, cows = 0;
        int[] count = new int[10];
        for(int i = 0; i < secret.length(); i++){
        	if(secret.charAt(i) == guess.charAt(i)){
        		bulls++;
        	}
        	else {
        		count[secret.charAt(i) - ‘0‘]++;
        		count[guess.charAt(i) - ‘0‘]--;
        		if(count[secret.charAt(i) - ‘0‘] <= 0)
        			cows++;
        		if(count[guess.charAt(i) - ‘0‘] >= 0)
        			cows++;
        	}
        }
        return bulls + "A" + cows + "B";
    }
}