給你一個字元串 time ,格式為 hh:mm(小時:分鐘),其中某幾位數字被隐藏(用 ? 表示)。
有效的時間為 00:00 到 23:59 之間的所有時間,包括 00:00 和 23:59 。
替換 time 中隐藏的數字,傳回你可以得到的最晚有效時間。
tips:考慮所有情況就好了
class Solution:
def maximumTime(self, time: str) -> str:
time = list(time)
if time[0] == '?' and time[1] != '?':
if int(time[1]) <= 3:
time[0] = '2'
else:
time[0] = '1'
elif time[0] == '?' and time[1] == '?':
time[0] = '2'
time[1] = '3'
if time[1] == '?' and time[0] == '0':
time[1] = '9'
elif time[1] == '?' and time[0] == '1':
time[1] = '9'
elif time[1] == '?' and time[0] == '2':
time[1] = '3'
if time[3] == '?':
time[3] = '5'
if time[4] == '?':
time[4] = '9'
return ''.join(time)