注意：凡是以英文出現(xiàn)的，都是題目提供的，包括答案代碼里的前幾行。
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.
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.
解析：
這一題本質(zhì)就是字符串比對(duì)，按理說(shuō)直接一次遍歷就可以搞定。但是題目要求cow重復(fù)的只算一個(gè)。也就是需要記錄是否重復(fù)。由于這里只有10個(gè)數(shù)字，因此可以用數(shù)組的元素值來(lái)記錄，開(kāi)辟10個(gè)int型空間即可。（如果字符的可能性很多，可以考慮用關(guān)聯(lián)容器來(lái)記錄，這里的數(shù)組也是用的關(guān)聯(lián)容器的思想，數(shù)組下標(biāo)當(dāng)作key，對(duì)應(yīng)的值當(dāng)作value）。arr[secret[i] - '0']++;
第二次遍歷判斷對(duì)應(yīng)的值之前是否存在過(guò)。arr[guess[j] - '0'] > 0
答案：
class Solution {
public:
string getHint(string secret, string guess) {
int arr[10] = { 0 };
int bull = 0;
int cow = 0;
for (int i = 0; i < secret.size(); i++) {
if (secret[i] == guess[i]) {
bull++;
} else {
arr[secret[i] - '0']++; //...
}
}
for (int j = 0; j < guess.size(); j++) {
if ((secret[j] != guess[j]) && (arr[guess[j] - '0'] > 0)) {
cow++;
arr[guess[j] - '0']--; //...
}
}
char res[10];
sprintf(res, "%dA%dB", bull, cow);
return res;
}
};
