【數學】C068_LC_日期之間隔幾天（枚舉 / 庫函數）

一、Problem

Write a program to count the number of days between two dates.

The two dates are given as strings, their format is YYYY-MM-DD as shown in the examples.

Input: date1 = "2020-01-15", date2 = "2019-12-31"
Output: 15

二、Solution

方法一：枚舉

• 直接算兩個日期之間隔的天數比較繁瑣，娶一個相對值，然後，算出兩個日期到這個相對值的各自天數 dd1 和 dd2…
• 最後返回 abs(dd1 - dd2) 即可…

class Solution {
    int[] days= {0,31,28,31,30,31,30,31,31,30,31,30,31};
    public int daysBetweenDates(String date1, String date2) {
        String[] ss1 = date1.split("-");
        String[] ss2 = date2.split("-");
        int y1 = Integer.parseInt(ss1[0]), y2 = Integer.parseInt(ss2[0]);
        int m1 = Integer.parseInt(ss1[1]), m2 = Integer.parseInt(ss2[1]);
        int d1 = Integer.parseInt(ss1[2]), d2 = Integer.parseInt(ss2[2]);
        return Math.abs(gap(y2, m2, d2) - gap(y1, m1, d1));
    }
    int gap(int yy, int mm, int dd) {
        for (int y = 1971; y < yy; y++) {
            dd += 365;
            if (isLeep(y))
                dd++;
        }
        for (int m = 1; m < mm; m++)
            dd += days[m];
        if (isLeep(yy) && mm > 2) dd++;	 //注
        return dd;
    }
    boolean isLeep(int y) {
        return y % 4 == 0 && y % 100 != 0 || y % 400 == 0;
    }
}

複雜度分析

• 時間複雜度：$O(...)$，
• 空間複雜度：$O(...)$，

---

方法二：庫函數

額…

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;
class Solution {
    public int daysBetweenDates(String date1, String date2) {
        SimpleDateFormat sdFormat = new SimpleDateFormat("yyyy-MM-dd");
        Date parse1 = null;
        Date parse2 = null;
        try {
            parse1 = sdFormat.parse(date1);
            parse2 = sdFormat.parse(date2);
        } catch(ParseException e) {
            e.printStackTrace();
        }
        long time = parse1.getTime();
        long time2 = parse2.getTime();
        long diff = time2 - time;
        return Math.abs((int)(diff / 1000 * 24 * 60 * 60));
    }
}

複雜度分析

• 時間複雜度：$O(...)$，
• 空間複雜度：$O(...)$，