LeetCode题目链接

注意：凡是以英文出现的，都是题目提供的，包括答案代码里的前几行。

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.
Credits:Special thanks to @jeantimex for adding this problem and creating all test cases.

解析：

这一题本质就是字符串比对，按理说直接一次遍历就可以搞定。但是题目要求cow重复的只算一个。也就是需要记录是否重复。由于这里只有10个数字，因此可以用数组的元素值来记录，开辟10个int型空间即可。（如果字符的可能性很多，可以考虑用关联容器来记录，这里的数组也是用的关联容器的思想，数组下标当作key，对应的值当作value）。arr[secret[i] - '0']++;
第二次遍历判断对应的值之前是否存在过。arr[guess[j] - '0'] > 0

答案：

class Solution {
public:
    string getHint(string secret, string guess) {
        int arr[10] = { 0 };
        int bull = 0;
        int cow = 0;
        
        for (int i = 0; i < secret.size(); i++) {
            if (secret[i] == guess[i]) {
                bull++;
            } else {
                arr[secret[i] - '0']++;    //...
            }
        }
        
        for (int j = 0; j < guess.size(); j++) {
            if ((secret[j] != guess[j]) && (arr[guess[j] - '0'] > 0)) {
                cow++;
                arr[guess[j] - '0']--;    //...
            }
        }
        
        char res[10];
        sprintf(res, "%dA%dB", bull, cow);
        return res;
    }
};