Leetcode题目：Bulls and Cows

Bulls and Cows 游戏提示算法

最新推荐文章于 2022-06-19 18:14:30 发布

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最新推荐文章于 2022-06-19 18:14:30 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/CodingGirl121/p/5418300.html

本文介绍了一种实现Bulls and Cows游戏提示的算法。该算法通过统计秘密数字与猜测数字中相同数字的位置和数量来提供提示。使用两个map结构分别记录秘密数字和猜测数字中的数字出现次数，并计算出正确位置的数字(A)和错误位置的数字(B)。

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.

题目解答：这个题目在实现的时候，代码写的比较复杂。首先，需要统计二者共同拥有的字母的个数。使用了两个map结构。应为B的数目中是不包括A的，还需要将对的数字在对的位置的这种情况减去。

代码如下：

class Solution {
public:
    string getHint(string secret, string guess) {
        map<char,int> B_map;
        map<char,int> B_map2;
        int A_count = 0;
        int B_count = 0;
        int len = secret.length();
        if((secret == "") || (guess == "") || (secret.length() != guess.length()))
            return "";
        for(int i = 0;i < len;i++)
        {
            if(B_map.find(secret[i]) == B_map.end())
                B_map[secret[i]] = 1;
            else
                B_map[secret[i]]++;
        }
        for(int i = 0;i < len;i++)
        {
            if(secret[i] == guess[i])
            {
                A_count++;
                B_map[secret[i]]--;
            }
            else
            {
                if(B_map2.find(guess[i]) == B_map2.end())
                    B_map2[guess[i]] = 1;
                else
                    B_map2[guess[i]]++;
            }
        }
        map<char,int>::iterator mit = B_map.begin();
        while(mit != B_map.end())
        {
            if(mit -> second > 0)
            {
                if(B_map2.find(mit -> first) != B_map2.end())
                {
                    int tmp =min(mit -> second , B_map2[mit -> first] );
                    B_count += tmp;
                }
            }
            mit++;
        }

        stringstream res;
        res.clear();
        res << A_count << 'A' << B_count <<'B';
        return res.str();
    }

    int min(int a, int b)
    {
        return a < b ? a : b;
    }
};