leetcode-1507-easy

Reformat Data

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.

思路一：分割字符，然后依次处理

    public String reformatDate(String date) {
        String[] arr = date.split("\\s");
        String day = arr[0];
        String month = arr[1];
        String year = arr[2];

        month = getMonth(month);
        day = getDay(day);

        return year + "-" + month + "-" + day;
    }

    private String getDay(String day) {
        if (day.length() == 3) {
            return "0" + day.charAt(0);
        } else {
            return day.substring(0, 2);
        }
    }

    private String getMonth(String month) {
        switch (month) {
            case "Jan":
                return "01";
            case "Feb":
                return "02";
            case "Mar":
                return "03";
            case "Apr":
                return "04";
            case "May":
                return "05";
            case "Jun":
                return "06";
            case "Jul":
                return "07";
            case "Aug":
                return "08";
            case "Sep":
                return "09";
            case "Oct":
                return "10";
            case "Nov":
                return "11";
            case "Dec":
                return "12";
            default:
                return "";
        }
    }