LeetCode(45)-Bulls and Cows

题目：

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.

思路：

• 题意：上面介绍的很清楚，就是给出两个整数字符串a和b，判断b中有多少个和a的中的整数相同，切位置相同，返回个数c，同时b中有多少和a中的整数相同，但是位置不同，个数是d，返回字符串“cAdB”。
• 字符串转化为数组，先判断c，遍历求相等的个数，对于求d，可以先把数组a的值以及值的重复个数存进hashMap叫做aa，然后把遍历b，把aa中存在的，村进去作为键，重复个数作为值，遍历相加这两个map的最小值，得到countB，countB-上面的个数c = d
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代码：

public class Solution {
    public String getHint(String secret, String guess) {
        char[] a = secret.toCharArray();
        char[] b = guess.toCharArray();
        int countA = 0;
        int countB = 0;
        //记录数组a的各元素的重复个数
        Map<Character,Integer> aa = new HashMap<Character,Integer>();
        //记录数组b的各元素的同时在a中出现的重复次数
        Map<Character,Integer> bb = new HashMap<Character,Integer>();
        for(int i = 0;i < a.length;i++){
            if(a[i] == b[i]){
                countA++;
            }
             if(aa.containsKey(a[i])){
                int f = aa.get(a[i]);
                f++;
                aa.put(a[i],f);
            }else{
                aa.put(a[i],1);
            }
        }
        for(int k = 0;k < a.length;k++){
            if(aa.containsKey(b[k])){
                if(bb.containsKey(b[k])){
                    int g = bb.get(b[k]);
                    g++;
                    bb.put(b[k],g);
                }else{
                    bb.put(b[k],1);
                }
            }
        }
        for(Character key:bb.keySet()){
            countB = countB+Math.min(aa.get(key),bb.get(key));
        }
        return countA+"A"+(countB-countA)+"B";
    }
}