LeetCode题解：Bulls and Cows

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.

题意：给定一个字符串表示一段密码，你的朋友给出他的猜测，你需要给他提示让他把密码猜出来。如果朋友猜的序列的某一位和密码对应位置的数相同，表示是A；如果有数字在密码序列中，但只是位置不对，则是B。最终返回一个序列表示有多少个A和多少个B。

解题思路：求A很简单，只要对应位置的字符相等则计数+1；对B计数的话，首先要知道有哪些序列是共有的，然后看有没有位置放错的，所以我们可以通过数组纪录每个数字出现的次数。

代码：

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