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.

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.

Please note that both secret number and friend's guess may contain duplicate digits.

Example 1:

Input: secret = "1807", guess = "7810"

Output: "1A3B"

Explanation: 1 bull and 3 cows. The bull is 8, the cows are 0, 1 and 7.

Example 2:

Input: secret = "1123", guess = "0111"

Output: "1A1B"

Explanation: The 1st 1 in friend's guess is a bull, the 2nd or 3rd 1 is a cow.

题意：

猜字游戏。

我负责猜，

你负责给提示：数字和位置都对，bulls++。数字对位置不对，cows++。

思路：

挨个扫字符串s，

挨个扫字符串g

若s当前字符等于g的字符，bulls++

用int[] map = new int[256]建一个简化版的HashMap

思路类似valid anagram

s当前字符在map里标记为++

p当前字符在map里标记为--

那么，若s当前字符已经被标记为--，说明p的字符来标记过，即它们字符相同但位置不同，cow++。

同理，若p当前字符已经被标记为++, 说明s的字符来标记过, 即它们字符相同但位置不同，cow++。

代码：

 class Solution {

     public String getHint(String secret, String guess) {

         // corner

         if(secret.length() != guess.length()) return false;

         int[] map = new int[256];

         int bull = 0;

         int cow = 0;

         for(int i = 0; i < secret.length();i++){

             char s = secret.charAt(i);

             char g = guess.charAt(i);

             if(s == g){

                 bull++;

             }else{

                 if(map[s]<0) cow++;

                 if(map[g]>0) cow++;

                 map[s]++;

                 map[g]--;

             }

         }

         return bull +"A" + cow + "B";

     }

 }