leetcode笔记：Bulls and Cows

一. 题目描写叙述

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.

二. 题目分析

该题即是猜数字（又称 Bulls and Cows ）是一种大概于20世纪中期兴起于英国的益智类小游戏。游戏规则抄送：http://baike.baidu.com/link?url=2OFEtJeMp3lxnGUt9tya4jVBwsUgCUM6C8F0zAdNTGtXS_KpjXa80lwIJpY_klUlL52JbO51fQYyqhs3fLMRLSsoSnGn7tvUOO6_iOxMTYbl7qEVhHdukYd1iTqqF-2uLDMXO9VuJFUTmmFN1qLFjVvQl6J6i7pr4Xd1oKJV-5_

在扫描过程前，创建一个数组用于记录0~9字符出现的次数。并在扫描时使用两种规则：

• 若字符在secret中出现一次则+1
• 若字符在guess中出现一次则-1

一个样例：当temp中’1’所相应的下标元素小于零时，此时说明guess此前出现过’1’的次数比secret多（或者说没有被secret出现’1’的次数抵消完），此时若secret出现’1’。则B++；反之，当temp中’1’所相应的下标元素大于零时。此时说明secret此前出现过’1’的次数比guess多（或者说没有被guess出现’1’的次数抵消完）。此时若guess出现’1’，则B++。

三. 演示样例代码

class Solution {
public:
    string getHint(string secret, string guess) {
        int temp[10] = {0}; // 用于存放0~9字符出现的次数
        int SIZE = secret.size();
        int A = 0, B = 0;
        for (int i = 0; i < SIZE; ++i)
        {
            if (secret[i] == guess[i])
            {
                ++A;
                continue;
            }
            else // 若字符在secret中出现一次则+1，在guess中出现一次则-1
            {
                if (temp[secret[i] - ‘0‘] < 0)
                    ++B;
                ++temp[secret[i] - ‘0‘];

                if (temp[guess[i] - ‘0‘] > 0)
                    ++B;
                --temp[guess[i] - ‘0‘];
            }
        }
        char result[10] = {0};  
        sprintf(result, "%dA%dB", A, B);
        return result;
    }
};

四. 小结

该题尽管比較简单，但还是挺有意思的，最后输出xAxB也改了一小段时间。