[leetcode] 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.

这道题是猜数游戏，题目难度为Easy。

根据题目的描述很容易想到用Hash Table来记录secret中每个数字出现的次数，然后依次统计bulls和cows的个数即可得出最终结果。由于只有10个数字，所以这里可以用vector来取代Hash Table记录次数。具体代码：

class Solution {
public:
    string getHint(string secret, string guess) {
        string hint = "";
        int bullCnt = 0, cowCnt = 0;
        vector<int> hash(10, 0);
        int sz = secret.size();
        
        for(int i=0; i<sz; ++i) hash[secret[i]-0x30]++;
        for(int i=0; i<sz; ++i) {
            if(secret[i] == guess[i]) {
                bullCnt++;
                hash[secret[i]-0x30]--;
            }
        }
        for(int i=0; i<sz; ++i) {
            if(secret[i] == guess[i]) continue;
            if(hash[guess[i]-0x30] > 0) {
                cowCnt++;
                hash[guess[i]-0x30]--;
            }
        }
        
        hint += to_string(bullCnt);
        hint += "A";
        hint += to_string(cowCnt);
        hint += "B";
        
        return hint;
    }
};

上面代码中字符串遍历了三次，看了别人的代码，只遍历一次即可得出最终结果，分享给大家：遍历secret和guess，同一位置数字不相等时，分别加减数字的计数，并以此判断当前数字是否是一个cow。具体代码：

class Solution {
public:
    string getHint(string secret, string guess) {
        string hint = "";
        int bullCnt = 0, cowCnt = 0;
        vector<int> cnt(10, 0);
        int sz = secret.size();
        
        for(int i=0; i<sz; ++i) {
            int sNum = secret[i] - 0x30;
            int gNum = guess[i] - 0x30;
            if(sNum == gNum) bullCnt++;
            else {
                if(cnt[sNum] > 0) cowCnt++;
                cnt[sNum]--;
                if(cnt[gNum] < 0) cowCnt++;
                cnt[gNum]++;
            }
        }
        
        hint += to_string(bullCnt);
        hint += "A";
        hint += to_string(cowCnt);
        hint += "B";
        
        return hint;
    }
​};