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.
​

解題

醜陋但是跑得快的解法:
Runtime: 0 ms, faster than 100.00%
Memory Usage: 2.1 MB, less than 38.71%
func reformatDate(date string) string {
d := strings.Fields(date)
ans := make([]byte, 0)
month := map[string]string{
"Jan": "01",
"Feb": "02",
"Mar": "03",
"Apr": "04",
"May": "05",
"Jun": "06",
"Jul": "07",
"Aug": "08",
"Sep": "09",
"Oct": "10",
"Nov": "11",
"Dec": "12",
}
for i:=0; i<len(d[2]); i++ {
ans = append(ans, d[2][i])
}
ans = append(ans, '-')
for i:=0; i<len(month[d[1]]); i++ {
ans = append(ans, month[d[1]][i])
}
ans = append(ans, '-')
if len(d[0])==3 {
ans = append(ans, '0')
}
for i:=0; i<len(d[0]); i++ {
if d[0][i]-'0'>=0 && d[0][i]-'0'<=9 {
ans = append(ans, d[0][i])
}
}
return string(ans)
}
​討論區看到的漂亮解法:
func reformatDate(date string) string {
month := map[string]string{
"Jan": "01",
"Feb": "02",
"Mar": "03",
"Apr": "04",
"May": "05",
"Jun": "06",
"Jul": "07",
"Aug": "08",
"Sep": "09",
"Oct": "10",
"Nov": "11",
"Dec": "12",
}
if len(date) == 12 { date = "0" + date }
return date[9:] + "-" + month[date[5:8]] + "-" + date[:2]
}
​