问题

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.

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.

例子

Secret number: "1807"
Friend's guess: "7810"

Secret number: "1123"
Friend's guess: "0111"

分析

设Secret number为S，Friend's guess为F.
bulls很容易统计出来，遍历一次就可以了，关键在于cows的处理。我们需要先遍历一次S和F，求出bulls，然后把S中没有被F匹配的字符的频率统计出来，再遍历一次S和F，即可求出cows。

要点

• two pass traverse;
• 键数量小的map可以用数组模拟，例如char.

时间复杂度

O(n)

空间复杂度

O(1)

代码

class Solution {
public:
    string getHint(string secret, string guess) {
        int count[10] = {0};
        int bulls = 0, cows = 0;
        for (int i = 0; i < guess.size(); i++) {
            if (guess[i] == secret[i]) bulls++;
            else count[secret[i] - '0']++;
        }
        for (int i = 0; i < guess.size(); i++) {
            if (guess[i] != secret[i] && count[guess[i] - '0'] > 0) {
                cows++;
                count[guess[i] - '0']--;
            }
        }
        return to_string(bulls) + "A" + to_string(cows) + "B";
    }
};