問題

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

設(shè)Secret number為S，F(xiàn)riend's guess為F.
bulls很容易統(tǒng)計(jì)出來，遍歷一次就可以了，關(guān)鍵在于cows的處理。我們需要先遍歷一次S和F，求出bulls，然后把S中沒有被F匹配的字符的頻率統(tǒng)計(jì)出來，再遍歷一次S和F，即可求出cows。

要點(diǎn)

• two pass traverse;
• 鍵數(shù)量小的map可以用數(shù)組模擬，例如char.

時(shí)間復(fù)雜度

O(n)

空間復(fù)雜度

O(1)

代碼

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";
    }
};