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)