leetcode [299]Bulls and Cows

Posted 2021-11-30 xiaobaituyun

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.

Note: You may assume that the secret number and your friend‘s guess only contain digits, and their lengths are always equal.

题目大意：

给定两个字符串，两个字符串只包含数字，而且两个字符串的长度相等。

解法：

使用一个数组来记录两个字符串中的字符，因为字符串中只含有数字，所以记录数组的范围就是0到9，如果说两个一样位置的字符相等，那么A直接加一。否则就使用nums数组进行记录，正数代表前面一个字符串出现过，而负数代表后面一个字符串出现过。当访问前面字符串中一个字符的nums小于0，说明该字符在后面的字符串出现过。反之也是同理。

java:

class Solution {
    public String getHint(String secret, String guess) {
        int A=0,B=0;
        int[] nums=new int[10];
        for (int i=0;i<secret.length();i++){
            int s=Character.getNumericValue(secret.charAt(i));
            int g=Character.getNumericValue(guess.charAt(i));
            if (s==g){
                A++;
            }else{
                if (nums[s]<0) B++;
                if (nums[g]>0) B++;
                nums[s]++;
                nums[g]--;
            }
        }
        return A+"A"+B+"B";
    }
}