Leetcode 299 Bulls and Cows

題目：
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.

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.

Please note that both secret number and friend’s guess may contain duplicate digits.

Example 1:

Input: secret = “1807”, guess = “7810”

Output: “1A3B”

Explanation: 1 bull and 3 cows. The bull is 8, the cows are 0, 1 and 7.
Example 2:

Input: secret = “1123”, guess = “0111”

Output: “1A1B”

Explanation: The 1st 1 in friend’s guess is a bull, the 2nd or 3rd 1 is a cow.
Note: You may assume that the secret number and your friend’s guess only contain digits, and their lengths are always equal.
這個題意挺簡單的，就是看對應位置上面兩個字串相同的個數與不在對應位置的數量，方法為使用一重for迴圈來計算重複的個數，最後的時候進行字串的拼接。
程式碼：
1)

class Solution {
public:
    string getHint(string secret, string guess) {
        int a = 0,b = 0,n = guess.size();
        int ds[10] = {0},dg[10] = {0};//標記相同的個數
        for(int i = 0 ; i < n ;i++){
            int x = secret[i] - '0';//將字串轉化為數字
            int y = guess[i] - '0';
            if(x == y){
                a++;
            }
            ds[x]++,dg[y]++;
        }
        for(int i = 0 ; i < 10 ; i++){
            b += min(ds[i],dg[i]);
        }
        b -= a;//除掉之後的數量
        return to_string(a) + 'A' + to_string(b) + 'B';
    }
};

2）

class Solution {
public:
    string getHint(string secret, string guess) {
        int bulls = 0, cows = 0;
        int cnt1[10], cnt2[10];
        memset(cnt1, 0, sizeof(cnt1));
        memset(cnt2, 0, sizeof(cnt2));
        for (int i = 0;i < secret.length(); i++) {
            if (guess[i] == secret[i]) {
                bulls++;
            } else {
                cnt1[secret[i] - '0'] ++;
                cnt2[guess[i] - '0'] ++;
            }
        }
        for (int i = 0;i < 10; i++) {
            cows += min(cnt1[i], cnt2[i]);
        }
        return to_string(bulls) + "A" + to_string(cows) + "B";
    }
};