Leetcode299. Bulls and Cows

题目：Leetcode299. Bulls and Cows                                    Difficulty: 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.

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.

Credits:

Special thanks to @jeantimex for adding this problem and creating all test cases.

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C++代码：1：为secret建立unordered_map,key对应数字，value对应出现次数。
                  2：第一次遍历，secret 与guess同位置相同，则countA++，hash[secret[i]]--   计算出A的值
                  3：第二次遍历，计算不同位置出现的相同数的个数
class Solution {
public:
string getHint(string secret, string guess) {
int countA=0;
int countB=0;
unordered_map<char,int> hash;
vector<bool> tag(secret.size(),false);
for(auto a: secret)
hash[a]++;
for(int i=0;i<secret.size();i++)
{
if(secret[i]==guess[i])
{
hash[secret[i]]--;
countA++;
tag[i]=true;

}
}
for(int i=0;i<guess.size();i++)
{
if(!tag[i]&&hash[guess[i]]>0)
{
countB++;
hash[guess[i]]--;

}
}
return to_string(countA)+"A"+to_string(countB)+"B";
}
};