299. Bulls and Cows | LeetCode

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.

beats 97%！我的思路如下，在循环时候做两个判断，第一个是判断是否有同位置的数字相等，相等则计入 bulls。第二个是判断是利用数字字符与数组下标可以一一对应的特点，以及 int 数组初始化时全为0的特点，用两个 int 数组来计算两个字符串里面出现的相同数字（不限位置）。
最后再来一次循环，统计两个字符的重复值，减去 bulls，即 cows。这样效率很高，都是数组读取。然而在讨论区看到了和我一样想法一样代码的人= =。。。。
代码如下：

public class Solution {    public String getHint(String secret, String guess) {        int len = secret.length();        int bulls = 0;        int cows = 0;        char [] sArr = secret.toCharArray();        char [] gArr = guess.toCharArray();        int [] sCount = new int[10];        int [] gCount = new int[10];        for(int i = 0; i < len; i++) {            if(sArr[i] == gArr[i]) {                bulls++;            }            sCount[sArr[i] - '0']++;            gCount[gArr[i] - '0']++;        }        for(int i = 0; i < 10; i++) {            cows += Math.min(sCount[i], gCount[i]);        }        cows -= bulls;        return bulls + "A" + cows + "B";    }}