zoukankan      html  css  js  c++  java
  • [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 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"
    

    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.

    思路:

    第一趟扫描查看有几个位置是正确的。统计一下。如果不正确分别放到map里面记录出现该数字的次数。

    最后再统计两个map里面的值,取最小的。

    string getHint(string secret, string guess)
    {
        if (secret.size() != guess.size() || guess.size() == 0)return "0A0B";
        int correct = 0, wrong = 0;
        map<char, int>se, gu;
        for (int i = 0; i < secret.size();i++)
        {
            if (secret[i] == guess[i])correct++;
            else
            {
                se[secret[i]]++;
                gu[guess[i]]++;
            }
        }
        for (char i = '0'; i <= '9'; i++)
        {
            wrong += min(se[i],wrong[i]);
        }
        return to_string(correct) + 'A' + to_string(wrong) + 'B';
    }

    参考:

    https://discuss.leetcode.com/topic/28445/c-4ms-straight-forward-solution-two-pass-o-n-time

  • 相关阅读:
    ABP 菜单 修改
    C# 过滤器
    RabbitMQ框架构建系列(三)——Net实现RabbitMQ之Producer
    RabbitMQ系列(二)RabbitMQ基础介绍
    RabbitMQ系列(一)AMPQ协议
    MVC 解读WebConfig
    MVC过滤器特性
    asp.net中使用JQueryEasyUI
    asp.net请求到响应的整个过程
    Redis的下载安装部署(Windows)
  • 原文地址:https://www.cnblogs.com/hellowooorld/p/7573387.html
Copyright © 2011-2022 走看看