leetcode oj java Bulls and Cows

一、问题描述：

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.

二、解决思路：

   bulls :  首先遍历找到对应位置相同的，如果相同bulls 加1；借助一个布尔型的二维数组，位置相同的地方置为false, 在计算cows的时候不参与计算。 时间为N

  cows : 因为是猜数字，用两个长度为10的二维数组来存放secrect 和guess中 每个数字出现的次数， 遍历两个数组每次取最小值叠加cows                      时间为3N

三、代码：

public String getHint(String secret, String guess) {         int bows = 0;        int cows = 0;        boolean[] flag = new boolean[secret.length()];        for (int i = 0; i < secret.length(); i++) {            flag[i] = true;            if (secret.charAt(i) == guess.charAt(i)) {                bows++;                flag[i] = false;            }        }        int[] sec = new int[10];        int[] gue = new int[10];        for (int i = 0; i < secret.length(); i++) {            if (flag[i]) {                char tmpsec = secret.charAt(i);                char tmpgue = guess.charAt(i);                sec[tmpsec - 48]++;                gue[tmpgue - 48]++;            }        }        for (int i = 0; i < 10; i++) {            cows += Math.min(sec[i], gue[i]);        }        return bows + "A" + cows + "B";    }

Tips : char  类型的可以直接用ascii 码值计算。