[leetcode299] 299. Bulls and Cows

public String getHint(String secret, String guess) {    /*    判断bull 是通过比较两个字符串的每一位，每次相同就删除该字符出现的次数，因为后边的    要判断cow不能占用这些    判断cow只要检测出去被占用的还有没有该字符     */    //key记录字符，value记录出现的次数    Map<Character,Integer> map = new HashMap<>();    String res = "";    int a = 0;    int b = 0;    boolean[] boo = new boolean[secret.length()];    for (int i = 0; i < secret.length(); i++) {        char cur = secret.charAt(i);        int num = map.getOrDefault(cur,0);        num++;        map.put(cur,num);    }    //判断bull和cow不能一起，要先判断bull，因为如果在判断cow的过程中会占用，会影响bull    for (int i = 0; i < guess.length(); i++) {        char cur = guess.charAt(i);        if (cur==secret.charAt(i))        {            a++;            boo[i] = true;            map.put(cur,map.get(cur)-1);        }    }    for (int i = 0; i < guess.length(); i++) {        char cur = guess.charAt(i);        if (!boo[i]&&map.containsKey(cur)&&map.get(cur)>0)        {            b++;            map.put(cur,map.get(cur)-1);        }    }    res = a+"A"+b+"B";    return res;}

原文地址：https://www.cnblogs.com/stAr-1/p/8323704.html