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.
答案：
A代表的是相同位置相同元素的个数，B记录的是错位的相同元素的个数。
思路是分两遍遍历，第一遍得到countA的个数，第二遍先建两个数组，因为两个字符串里面都只有数字，所以建两个大小为10的数组分别记录Secret和Guess中0到9的个数，然后再比较这两个数组，相同位置上小的那个数被记录下来，然后再累加。
1 class Solution {
2 public:
3 string getHint(string secret, string guess) {
4 int countA=0,countB=0;
5 int count1[10];
6 int count2[10];
7 memset(count1,0,sizeof(count1));
8 memset(count2,0,sizeof(count2));
9 for(int i=0;i<secret.length();i++){
10 if(secret[i]==guess[i]){
11 countA++;
12 }
13 else{
14 count1[secret[i]-‘0‘]++;
15 count2[guess[i]-‘0‘]++;
16 }
17 }
18 for(int i=0;i<10;i++){
19 countB=countB+min(count1[i],count2[i]);
20 }
21 return to_string(countA)+"A"+to_string(countB)+"B";
22 }
23 };