LeetCode题目链接,二. 题目分析

作者: 编程  发布:2019-08-28

leetcode笔记:Bulls and Cows

一. 主题材料陈说

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. 标题剖判

该题正是猜数字(又称 Bulls and Cows )是一种大约于20世纪中期兴起于United Kingdom的益智类小游戏。游戏法规抄送:

在围观进程前,创制三个数组用于记录0~9字符出现的次数,并在扫描时选择二种准绳:

若字符在secret中出现二回则 1 若字符在guess中冒出一次则-1

一个例证:当temp中’1’所对应的下标成分小于零时,此时证实guess在此以前边世过’1’的次数比secret多(恐怕说未有被secret出现’1’的次数抵消完),此时若secret现身’1’,则B ;反之,当temp中’1’所对应的下标成分大于零时,此时申明secret在此以前出现过’1’的次数比guess多(或然说未有被guess出现’1’的次数抵消完),此时若guess出现’1’,则B 。

三. 示例代码

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, "
					

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