1507. Reformat Date
Given a date
string in the form Day Month Year
, where:
Day
is in the set{"1st", "2nd", "3rd", "4th", ..., "30th", "31st"}
.Month
is in the set{"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"}
.Year
is in the range[1900, 2100]
.
Convert the date string to the format YYYY-MM-DD
, where:
YYYY
denotes the 4 digit year.MM
denotes the 2 digit month.DD
denotes the 2 digit day.
Example 1:
Input: date = "20th Oct 2052" Output: "2052-10-20"
Example 2:
Input: date = "6th Jun 1933" Output: "1933-06-06"
Example 3:
Input: date = "26th May 1960" Output: "1960-05-26"
Constraints:
- The given dates are guaranteed to be valid, so no error handling is necessary.
class Solution { public String reformatDate(String date) { //if(date.equals("") || date == null) return ""; String[] dat = date.split(" "); Map<String, String> map = new HashMap(); map.put("Jan", "01"); map.put("Feb","02"); map.put("Mar","03"); map.put("Apr","04"); map.put("May","05"); map.put("Jun","06"); map.put("Jul","07"); map.put("Aug","08"); map.put("Sep","09"); map.put("Oct","10"); map.put("Nov","11"); map.put("Dec","12"); StringBuilder sb = new StringBuilder(); char[] day = dat[0].toCharArray(); String mon = dat[1]; String yea = dat[2]; sb.append(yea).append("-"); sb.append(map.get(mon)).append("-"); int i = 0; int cur = 0; while(Character.isDigit(day[i])){ cur = 10 * cur + (day[i] - '0'); i++; } if(cur < 10) sb.append(0); sb.append(cur); return sb.toString(); } }
要注意的就是日子前面的0和Character.isDigit(char c)的用法
总结:
nothing to worry about year.
for month we use a string map
for days we need to take care of 0-9 (add one more 0 infront of days)