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  • 0299. Bulls and Cows (E)

    Bulls and Cows (E)

    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.

    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.


    题意

    猜数字游戏,返回"aAbB"形式的反馈信息。

    思路

    两次循环:第一次记录secret中所有数字的个数,第二次找位置正确的bulls同时,找guess数字在secret中出现的个数total,total - bulls就是cows。

    一次循环:如果对应字符相等,则bulls++;如果不相等,若secret对应字符出现次数为负数,说明该字符之前在guess中出现过,则cows++,若guess对应字符出现次数为正数,说明该字符之前在secret中出现过,则cows++。


    代码实现

    Java

    两次循环

    class Solution {
        public String getHint(String secret, String guess) {
            int[] nums = new int[10];
            for (char c : secret.toCharArray()) {
                nums[c - '0']++;
            }
    
            int A = 0, total = 0;
            for (int i = 0; i < secret.length(); i++) {
                if (secret.charAt(i) == guess.charAt(i)) {
                    A++;
                }
                if (nums[guess.charAt(i) - '0'] != 0) {
                    nums[guess.charAt(i) - '0']--;
                    total++;
                }
            }
    
            return A + "A" + (total - A) + "B";
        }
    }
    

    一次循环

    class Solution {
        public String getHint(String secret, String guess) {
            int nums[] = new int[10];
            int A = 0, B = 0;
    
            for (int i = 0; i < secret.length(); i++) {
                char s = secret.charAt(i);
                char g = guess.charAt(i);
    
                if (s == g) {
                    A++;
                } else {
                    if (nums[s - '0'] < 0) B++;
                    if (nums[g - '0'] > 0) B++;
                    nums[s - '0']++;
                    nums[g - '0']--;
                }
            }
    
            return A + "A" + B + "B";
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/mapoos/p/13155702.html
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