1507. Reformat Date ---------------------E

Q:

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.

A:

class Solution {
public:
    string reformatDate(string date) {
        istringstream iss(date);
        vector<string> str( (istream_iterator<string>(iss)), istream_iterator<string>());
        string Day = str[0];
        string Month = str[1];
        string Year = str[2];
        
        string DD = Day.substr(0,Day.length()-2);
        if(DD.length()==1)
            DD = "0"+DD;
        
        vector<string> V{"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};     
        string MM = "";
        for(int i = 0;i<V.size(); ++i){
            if(V[i]==Month)
            {
                MM = to_string(i+1);
                if(MM.length()<2)
                    MM = "0"+MM;
            }
        }
        return Year+"-"+MM+"-"+DD;
    }
};