[leetcode] 299. Bulls and Cows

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本文介绍了一种计算Bulls和Cows游戏提示的方法。通过两次遍历字符串，分别计算bulls（位置和数字都匹配的情况）和cows（仅数字匹配但位置不对的情况）。提供了两种解法，包括如何处理重复数字。

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.

解法一：

第一次遍历计算bull的个数，并记录secret中不是bull的数字出现频率。第二次遍历看guess中不是bull但在secret中出现的数字，并记录次数。

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';
        
    }
};

解法二：

我们看如果secret当前位置数字的映射值小于0，则表示其在guess中出现过，cows自增1，然后映射值加1，如果guess当前位置的数字的映射值大于0，则表示其在secret中出现过，cows自增1，然后映射值减1。

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";
    }
};