题目：

You are playing the following 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;

    }

};