[LeetCode] Bulls and Cows 公母牛游戲

 

You are playing the following Bulls and Cows game with your friend: You write a 4-digit secret number and ask your friend to guess it, each time your friend guesses a number, you give a hint, the hint tells your friend how many digits are in the correct positions (called "bulls") and how many digits are in the wrong positions (called "cows"), your friend will use those hints to find out 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.)

 

According to Wikipedia: "Bulls and Cows (also known as Cows and Bulls or Pigs and Bulls or Bulls and Cleots) is an old code-breaking mind or paper and pencil game for two or more players, predating the similar commercially marketed board game Mastermind. The numerical version of the game is usually played with 4 digits, but can also be played with 3 or any other number of digits."

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.

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.

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這道題提出了一個叫公牛母牛的游戲，其實就是之前文曲星上有的猜數字的游戲，有一個四位數字，你猜一個結果，然后根據你猜的結果和真實結果做對比，提示有多少個數字和位置都正確的叫做bulls，還提示有多少數字正確但位置不對的叫做cows，根據這些信息來引導我們繼續猜測正確的數字。這道題並沒有讓我們實現整個游戲，而只用實現一次比較即可。給出兩個字符串，讓我們找出分別幾個bulls和cows。這題需要用哈希表，來建立數字和其出現次數的映射。我最開始想的方法是用兩次遍歷，第一次遍歷找出所有位置相同且值相同的數字，即bulls，並且記錄secret中不是bulls的數字出現的次數。然后第二次遍歷我們針對guess中不是bulls的位置，如果在哈希表中存在，cows自增1，然后映射值減1，參見如下代碼：

 

解法一：

class Solution {
public:
    string getHint(string secret, string guess) {
        int m[256] = {0}, bulls = 0, cows = 0;
        for (int i = 0; i < secret.size(); ++i) {
            if (secret[i] == guess[i]) ++bulls;
            else ++m[secret[i]];
        }
        for (int i = 0; i < secret.size(); ++i) {
            if (secret[i] != guess[i] && m[guess[i]]) {
                ++cows;
                --m[guess[i]];
            }
        }
        return to_string(bulls) + "A" + to_string(cows) + "B";
    }
};

 

我們其實可以用一次循環就搞定的，在處理不是bulls的位置時，我們看如果secret當前位置數字的映射值小於0，則表示其在guess中出現過，cows自增1，然后映射值加1，如果guess當前位置的數字的映射值大於0，則表示其在secret中出現過，cows自增1，然后映射值減1，參見代碼如下：

解法二：

class Solution {
public:
    string getHint(string secret, string guess) {
        int m[256] = {0}, bulls = 0, cows = 0;
        for (int i = 0; i < secret.size(); ++i) {
            if (secret[i] == guess[i]) ++bulls;
            else {
                if (m[secret[i]]++ < 0) ++cows;
                if (m[guess[i]]-- > 0) ++ cows;
            }
        }
        return to_string(bulls) + "A" + to_string(cows) + "B";
    }
};

 

最后我們還可以稍作修改寫的更簡潔一些，a是bulls的值，b是bulls和cows之和，參見代碼如下：

解法三：

class Solution {
public:
    string getHint(string secret, string guess) {
        int m[256] = {0}, a = 0, b = 0, i = 0;
        for (char s : secret) {
            char g = guess[i++];
            a += s == g;
            b += (m[s]++ < 0) + (m[g]-- > 0);
        }
        return to_string(a) + "A" + to_string(b - a) + "B";
    }
};

 

參考資料：

https://leetcode.com/discuss/67031/one-pass-java-solution

https://leetcode.com/discuss/67125/short-c-o-n

https://leetcode.com/discuss/67012/c-one-pass-o-n-time-o-1-space

 

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