[Leetcode]299. Bulls and Cows

题目描述

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.

例子

例子 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.

例子 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.

其他

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

解题思路

简单来说，给定两个相等长度，只包括 '0' ~ '9' 的字符串；要求找到位置相同且相等的数字数量，和有共同数字但是位置不同的数量；考虑到要统计出现过的数字的数量，应该要用哈希表。思路可以是先通过哈希表记录两个字符串里出现过次数的统计；然后通过一次遍历去除掉位置也相同的数字数量，然后再比较两个哈希表，对每一个数字取两个字符串中包含数量的最小值就可以，代码如下：

class Solution {
public:
    string getHint(string secret, string guess) {
        vector<int> digit_nums_secret(10, 0);
        vector<int> digit_nums_guess(10, 0);
        for (char digit: secret) {
            digit_nums_secret[digit - '0']++;
        }

        for (char digit: guess) {
            digit_nums_guess[digit - '0']++;
        }

        string hint = "";
        int num_bulls = 0, num_cows = 0;

        for (int i = 0; i < secret.length(); i++) {
            if (secret[i] == guess[i]) {
                num_bulls++;
                int index = secret[i] - '0';
                digit_nums_secret[index]--;
                digit_nums_guess[index]--;
            }
        }

        hint += (to_string(num_bulls) + "A");

        for (int i = 0; i < 10; i++) {
            num_cows += min(digit_nums_secret[i], digit_nums_guess[i]);
        }

        hint += (to_string(num_cows) + "B");

        return hint;
    }
};

时间复杂度：O(n)
空间复杂度：O(n)