第十七周leetcode题

Description:

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.

对于给定的两串数secret和guess，相同位置且相等的数称为bull，不同位置且相等的数称为cow。因为bull对应的两个数和cow对应的两个数分别相等，因此可以先找两串数中相等的数。分别统计两串数中每个数字出现的次数。再同时检索两个数串，相同位置的数值相等的话是bull，每发现一次bull就在统计好的两串数中每个数字出现的次数里将bull对应的数的次数减一，减完所有bull后，再按数字从小到大的顺序同时检索两个数串中每个数字出现的次数，取两者中的较小值加进cow中（因为较小值即为除去bull后数值相等的数对的个数）即为所求。代码如下：

class Solution {
public:
    string getHint(string secret, string guess) {
        int a[10]={};
        int b[10]={};
        int bull=0;
        int cow=0;
        
       
        
        for(int i=0;i<secret.length();i++)
        {
            a[secret[i]-'0']++;
            b[guess[i]-'0']++;           
        }
        
        for(int i=0;i<secret.length();i++)
        {
            if(secret[i]==guess[i])
            {
                bull++;
                a[secret[i]-'0']--;
                b[guess[i]-'0']--;  
            }
          
        }
        
        for(int i=0;i<10;i++)
        {
            cow+=min(a[i],b[i]);
        }
        
        
        
        return  to_string(bull)+'A'+to_string(cow)+'B';
        
        
    }
};