# [leetcode]299. Bulls and Cows-猜数字

### 题目和简单翻译

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.

Secret number:  "1807"
Friend's guess: "7810"


1 个 bull, 3 cows. (The bull is 8, the cows are 01 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"


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.

### 解题思路

1. 大家都能想到的方法, 做两个向量, 分别存谜底和猜测结果. 两个向量有重复索引值不为 0, 则 cows++;
class Solution {
public:
string getHint(string secret, string guess) {
int resoultA = 0;
int resoultB = 0;
vector<int> hintA(10,0);
vector<int> hintB(10,0);
for(int i=0;i<secret.length();i++){
if(secret[i]==guess[i]) resoultA++;
else {
hintA[secret[i]-'0']++;
hintB[guess[i]-'0']++;
}
}
for(int i=0;i<10;i++){
cout<<hintA[i]<<" "<<hintB[i]<<",";
resoultB += min(hintA[i],hintB[i]);
}
}
};

1. 发现的一个十分聪明的算法, 只需要一个向量, 谜底触发 +1, 猜测结果触发 -1 都触发结果抵消, 此时 cow++, 整个过程中, 只循环触发过的数字, 省时.
class Solution {
public:
string getHint(string secret, string guess) {
vector<int> num(10,0);
int bull = 0, cow = 0;
for (unsigned i = 0; i < secret.size(); ++i) {
if (secret[i] == guess[i]) ++bull;
else {
num[secret[i] - '0']++;
num[guess[i] - '0']--;
if (num[guess[i] - '0'] >= 0) ++cow;
if ( num[secret[i] - '0'] <= 0) ++cow;
}
}