一天一道LeetCode

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（一）题目

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.

来源： https://leetcode.com/problems/bulls-and-cows/

（二）解题

题目大意：给定两个string变量a和b，bulls表示a和b相同位上有相同数的个数，cows表示a和b不同位上有相同数的个数。注意计算过的不能再算。

解题思路：分为两类来计算，如1807和7810

(1)相同位上有相同数(8,8)

直接统计这样的位数即可

(2)相同位上有不同数(107和710)

先统计每个数出现

具体思路见代码注释：

class Solution {
public:
    string getHint(string secret, string guess) {
        int hash[10] = {0};//0~9在secret出现的次数
        int len = secret.length();
        int countA = 0 ,countB=0;
        int *isFind = new int[len];//统计那些位上的数相同
        memset(isFind,0,len*sizeof(int));
        for(int i =0 ; i < len ;i++)
        {
            if(secret[i]==guess[i]){//相同位上有相同数
                isFind[i] = 1;
                countA++;
            }
            else hash[secret[i]-'0']++;//如果相同位上数不相等就记录下来
        }
        for(int i =0 ; i < len ;i++)
        {
            if(isFind[i]==0){//跳过相同位上有相同数
                if(hash[guess[i]-'0']!=0)//如果出现过
                {
                    hash[guess[i]-'0']--;//次数减1
                    countB++;
                }
            }
        }
        string ret;
        stringstream ss;
        ss<<countA;
        ss<<'A';
        ss<<countB;
        ss<<'B';
        ss>>ret;//按特定格式输出
        return ret;
    }
};