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Formatted question description: https://leetcode.ca/all/1507.html

1507. Reformat Date (Easy)

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.

Related Topics:
String

Solution 1.

// OJ: https://leetcode.com/problems/reformat-date/
// Time: O(1)
// Space: O(1)
class Solution {
public:
    string reformatDate(string date) {
        istringstream ss(date);
        string day, mon, year, m;
        ss >> day >> mon >> year;
        if (mon == "Jan") m = "01";
        else if (mon == "Feb") m = "02";
        else if (mon == "Mar") m = "03";
        else if (mon == "Apr") m = "04";
        else if (mon == "May") m = "05";
        else if (mon == "Jun") m = "06";
        else if (mon == "Jul") m = "07";
        else if (mon == "Aug") m = "08";
        else if (mon == "Sep") m = "09";
        else if (mon == "Oct") m = "10";
        else if (mon == "Nov") m = "11";
        else m = "12";
        string d = to_string(stoi(day));
        return year + "-" + m + "-" + (d.size() == 1? "0" + d : d);
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/reformat-date/
// Time: O(1)
// Space: O(1) 
class Solution {
public:
    string reformatDate(string date) {
        istringstream ss(date);
        string day, mon, year, m;
        ss >> day >> mon >> year;
        string months[12] = {"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec" };
        for (int i = 0; i < 12; ++i) {
            if (months[i] == mon) {
                m = to_string(i + 1);
                if (m.size() == 1) m = "0" + m;
                break;
            }
        }
        string d = to_string(stoi(day));
        return year + "-" + m + "-" + (d.size() == 1? "0" + d : d);
    }
};

Java

  • class Solution {
    public String reformatDate(String date) {
        String[] months = {"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};
        Map<String, Integer> map = new HashMap<String, Integer>();
        for (int i = 1; i <= 12; i++)
            map.put(months[i - 1], i);
        String[] array = date.split(" ");
        int year = Integer.parseInt(array[2]);
        int month = map.get(array[1]);
        int day = Integer.parseInt(array[0].substring(0, array[0].length() - 2));
        return String.format("%d-%02d-%02d", year, month, day);
    }
}

  • // OJ: https://leetcode.com/problems/reformat-date/
// Time: O(1)
// Space: O(1)
class Solution {
public:
    string reformatDate(string date) {
        istringstream ss(date);
        string day, mon, year, m;
        ss >> day >> mon >> year;
        if (mon == "Jan") m = "01";
        else if (mon == "Feb") m = "02";
        else if (mon == "Mar") m = "03";
        else if (mon == "Apr") m = "04";
        else if (mon == "May") m = "05";
        else if (mon == "Jun") m = "06";
        else if (mon == "Jul") m = "07";
        else if (mon == "Aug") m = "08";
        else if (mon == "Sep") m = "09";
        else if (mon == "Oct") m = "10";
        else if (mon == "Nov") m = "11";
        else m = "12";
        string d = to_string(stoi(day));
        return year + "-" + m + "-" + (d.size() == 1? "0" + d : d);
    }
};

  • class Solution:
    def reformatDate(self, date: str) -> str:
        s = date.split()
        s.reverse()
        months = " JanFebMarAprMayJunJulAugSepOctNovDec"
        s[1] = str(months.index(s[1]) // 3 + 1).zfill(2)
        s[2] = s[2][:-2].zfill(2)
        return "-".join(s)

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