算法－第四版－练习1.2.12解答

为SmartDate添加一个方法dayOfTheWeek()，为日期中每周的日返回Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday中的适当值。你可以假定时间是21世纪。

/**
 * Description : 
 * Author      : mn@furzoom.com
 * Date        : Sep 27, 2016 9:45:00 AM
 * Copyright (c) 2013-2016, http://furzoom.com All Rights Reserved.
 */
package com.furzoom.lab.algs.ch102;

class SmartDate
{
    private final int month;
    private final int day;
    private final int year;
    private final int[] months = 
        {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    private final int[] days = 
        {0, 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334};
    private final int[] daysLeap =  
        {0, 0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335};
    private final String[] weekdays = {"Sunday", "Monday", "Tuesday", "Wednesday", 
            "Thursday", "Friday", "Saturday"};
    
    public SmartDate(int m, int d, int y)
    {
        if (!validate(m, d, y))
            throw new IllegalArgumentException("Argument illegal " + m + "/" + d + "/" + y);
        this.month = m;
        this.day = d;
        this.year = y;
    }
    
    public int month()
    {
        return month;
    }
    
    public int day()
    {
        return day;
    }
    
    public int year()
    {
        return year;
    }
    
    public String dayOfTheWeek()
    {
        // based on 1/1/2000
        int totalDays;
        if (isLeapYear())
        {
            totalDays = daysLeap[month] + day;
        }
        else
        {
            totalDays = days[month] + day;
        }
        
        for (int i = 2000; i < year; i++)
        {
            if (isLeapYear(i))
                totalDays += 366;
            else
                totalDays += 365;
        }
        
        // 1/1/2000 is Saturday
        return weekdays[((totalDays - 1) % 7 + 6) % 7];
    }
    
    public String toString()
    {
        return month + "/" + day + "/" + year;
    }
    
    private boolean validate(int m, int d, int y)
    {
        if (y == 0 || y < -1000 || y > 10000)
            return false;
        
        if (m < 1 || m > 12)
            return false;
        
        if (d < 1 || d > 31)
            return false;
        
        if (d > months[m])
            return false;
        
        if (!isLeapYear() && d > 28)
            return false;
        
        return true;
    }
    
    private boolean isLeapYear(int y)
    {
        if (y % 400 == 0 || (y % 4 == 0 && y % 100 != 0))
        {
            return true;
        }
        else
        {
            return false;
        }
    }
    
    private boolean isLeapYear()
    {
        return isLeapYear(year);
    }
}

测试：

/**
 * Description : 
 * Author      : mn@furzoom.com
 * Date        : Sep 27, 2016 9:42:39 AM
 * Copyright (c) 2013-2016, http://furzoom.com All Rights Reserved.
 */
package com.furzoom.lab.algs.ch102;

/**
 * ClassName    : E10212 <br>
 * Function     : TODO ADD FUNCTION. <br>
 * date         : Sep 27, 2016 9:42:39 AM <br>
 * 
 * @version 
 */
public class E10212
{
    public static void main(String[] args)
    {
        SmartDate date = new SmartDate(1, 1, 2000);
        System.out.println(date + " " + date.dayOfTheWeek());
        
        date = new SmartDate(2, 1, 2000);
        System.out.println(date + " " + date.dayOfTheWeek());
        
        date = new SmartDate(12, 29, 2000);
        System.out.println(date + " " + date.dayOfTheWeek());
        
        date = new SmartDate(12, 29, 2010);
        System.out.println(date + " " + date.dayOfTheWeek());
    }

}

结果：

1/1/2000 Saturday
2/1/2000 Tuesday
12/29/2000 Friday
12/29/2010 Wednesday

算法－第四版－1.2 数据抽象－习题索引汇总

算法－第四版习题索引汇总