(Easy) Day of the week - LeetCode
时间:2019-09-09
本文章向大家介绍(Easy) Day of the week - LeetCode,主要包括(Easy) Day of the week - LeetCode使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。
Description:
Given a date, return the corresponding day of the week for that date.
The input is given as three integers representing the day
, month
and year
respectively.
Return the answer as one of the following values {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}
.
Example 1:
Input: day = 31, month = 8, year = 2019 Output: "Saturday"
Example 2:
Input: day = 18, month = 7, year = 1999 Output: "Sunday"
Example 3:
Input: day = 15, month = 8, year = 1993 Output: "Sunday"
Constraints:
- The given dates are valid dates between the years
1971
and2100
.
Accepted
4,400
Submissions
6,472
Solution:
class Solution { public String dayOfTheWeek(int day, int month, int year) { //2019-9-8 Sunday. // we need to get the date difference. In that way, we will be able to know what day that day is. int num_1 = GetNum(day,month,year); int num_2 = GetNum(8,9,2019); //System.out.println(num_2-num_1); int diff = Math.abs(num_2-num_1)%7; if(diff ==0){ return "Sunday"; } if(num_2 - num_1 >0){ if(diff==1){ return "Saturday"; } else if(diff == 2){ return "Friday"; } else if(diff == 3){ return "Thursday"; } else if(diff == 4){ return "Wednesday"; } else if(diff == 5){ return "Tuesday"; } else if(diff == 6){ return "Monday"; } } else{ if(diff==1){ return "Monday"; } else if(diff == 2){ return "Tuesday"; } else if(diff == 3){ return "Wednesday"; } else if(diff == 4){ return "Thursday"; } else if(diff == 5){ return "Friday"; } else if(diff == 6){ return "Saturday"; } } System.out.println(GetNum(3,3,1977)); System.out.println(GetNum(8,9,2019)); return ""; } /* 给定一个年份,判断是否是闰年。条件为: A:能被4整除,并且不能被100整除。或者 B:能被400整除。*/ static boolean Leap(int a){ if(a%4==0&& a%100!=0){ return true; } else if( a%400 ==0){ return true; } else{ return false; } } static int GetNum(int day, int month, int year){ HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); //Jan 31 days //Feb Leap year or not. (28/29 days) //Mar 31 days //Apr 30 days //May 31 days //Jun 30 days //Jul 31 days //Aug 31 days //Sep 30 days //Oct 31 days //Nov 30 days //Dec 31 days map.put(1,31); map.put(3,31); map.put(4,30); map.put(5,31); map.put(6,30); map.put(7,31); map.put(8,31); map.put(9,30); map.put(10,31); map.put(11,30); map.put(12,31); if(Leap(year)){ map.put(2,29); } else{ map.put(2,28); } int sum = 0; for(int i = 1977; i< year; i++){ if(Leap(i)){ sum = sum + 366; } else{ sum = sum +365; } } for(int i = 1; i < month; i++){ sum = sum + map.get(i); } sum = sum + day; return sum; } }
原文地址:https://www.cnblogs.com/codingyangmao/p/11490892.html
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