Leetcode @ Bulls and Cows

299. Bulls and Cows

Easy

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.

理解题意很重要，里面只算digit，而不是两个连在一起的数字，比如11 vs 110，算2bulls，而不是1 bull，并不是把"11"作为一个数字去看，最开始我是这样想的就想复杂了。

理解清楚题意后，就变得很简单，用两个数组来分别存当secret 和 guess某一位 数字不想等时的count，如果相等，就bull++，最后在iterate through 两个数组，大小为10，取每位上两者中的最小的那个，因为要找重合的个数么，所以找交集。

class Solution {
public:
    string getHint(string secret, string guess) {
        int size = secret.size();
        vector<int> rec1(10, 0), rec2(10, 0);
        int bulls = 0, cows = 0;
        for (int i = 0; i<size; i++)
        {
            if (secret[i] == guess[i])
            {
                bulls++;
            }
            else
            {
                rec1[secret[i]-'0']++;
                rec2[guess[i]-'0']++;
            }
        }
        for (int i = 0; i<=9; i++)
        {
            cows += min(rec1[i], rec2[i]);
        }
        return to_string(bulls) + "A" + to_string(cows) + "B";
    }
};