leetcode_ Bulls and Cows

以下为问题描述：
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.

其实这就是我们小时候上学时候玩过的猜数游戏，小A心里想一个随机数让小B猜，根据小B的猜测给出相应的提示，直到小B猜对为止。我们要写的这个算法的功能就是要根据小A和小B的数字来计算出正确的提示。

思路：先构建guess和secret的两个哈希表，并统计出A（位置和数字正确）的数量，再将两个哈希表里都有的项取最小值再减去A得出B（数字正确但位置错误）的数量。

代码如下

class Solution {public:    string getHint(string secret, string guess) {        string result;        int ACount = 0;                             //A的数量        int BCount = 0;                             //B的数量        vector<int> secretHash(10,0);               //用来存放随机数各位数字的哈希表        vector<int> guessHash(10,0);                //用来存放猜测数各位数字的哈希表        //统计A的数量        for (int i = 0; i < secret.length(); ++i)        {            secretHash[secret[i]-'0']++;            guessHash[guess[i]-'0']++;            if (secret[i] == guess[i])            {                ++ACount;                                       }        }        //累加疑似B的数量        //如1122和0001这对输入中secretHash[1]=2,guessHash[1]=1，B的疑似值为1(min(2,1))        for (int i = 0; i < secretHash.size(); ++i)        {            if(guessHash[i] > 0 && secretHash[i] > 0)            {                BCount += min(secretHash[i],guessHash[i]);            }        }        BCount -= ACount;        if (BCount < 0)        {            BCount = 0;        }        char buf[1024];        sprintf(buf,"%dA%dB",ACount,BCount);        result = buf;        return result;    }};