299. Bulls and Cows

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.

题意：给出两个字符串，找出有几个同位置相等，有几个相等但不同位置。

思路：首先遍历一遍，把同位置相等的bull数找出来，并把不等的存入unordered_multiset。第二遍找cow的数。分两遍的原因是bull数的优先级比cow数的优先级高，如"1122"&&"1222"。注意：unordered_multiset的删除的使用。

class Solution {
public:
string getHint(string secret, string guess) {
unordered_multiset<char> myset;

int bull = 0;
int cow = 0;
for (int i = 0; i < secret.size(); i++){
if (secret[i] == guess[i])
bull++;
else{
myset.insert(secret[i]);
}
}
for (int i = 0; i < secret.size(); i++){
if (secret[i] != guess[i]){
unordered_multiset<char>::iterator it = myset.find(guess[i]);
if (it != myset.end()){
cow++;
myset.erase(it);
}
}
}
stringstream ss;
ss << bull << "A" << cow << "B";
return ss.str();
}
};