Bulls and Cows

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 are0, 1 and 7.)

Write a function to return a hint according to the secret number and friend's guess, useA 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".

分析：

       一次遍历找出位置相同的，并记录两个字符串位置不同的字符出现次数，那么数字相同位置不同的数就是每次两个字符串中出现字符次数较小的数。

代码：

class Solution {public:    string getHint(string secret, string guess) {        int seclength=secret.length();        int guelength=guess.length();        int sameA=0;        int difB=0;        int sectable[11];        int guesstable[11];        memset(sectable,0,sizeof(sectable));        memset(guesstable,0,sizeof(guesstable));        for(int i=0; i<seclength; i++)        {            if(secret[i]==guess[i])            {                sameA++;            }            else            {                sectable[secret[i]-'0']++;                guesstable[guess[i]-'0']++;            }                                }        for(int i=0; i<10; i++)        {            difB+=(sectable[i]<guesstable[i]?sectable[i]:guesstable[i]);        }        stringstream ss;        ss<<sameA;         string s1 = ss.str();         stringstream ss1;        ss1<<difB;        string s2 =ss1.str();        return s1+'A'+s2+'B';            }};