[LeetCode]Next Closest Time

题目描述:

LeetCode 681. Next Closest Time

Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused.

You may assume the given input string is always valid. For example, "01:34", "12:09" are all valid. "1:34", "12:9" are all invalid.

Example 1:

Input: "19:34"
Output: "19:39"
Explanation: The next closest time choosing from digits 1, 9, 3, 4, is 19:39, which occurs 5 minutes later.  It is not 19:33, because this occurs 23 hours and 59 minutes later.

Example 2:

Input: "23:59"
Output: "22:22"
Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22. It may be assumed that the returned time is next day's time since it is smaller than the input time numerically.

题目大意:

给定格式为HH:MM的时间time,根据其中的数字,组成距离当前时间最近的下一个时刻。

解题思路:

将time的数字取出并排序,记为stime,令X = stime[0]

从time的低位向高位枚举,将stime中恰好大于当前值的值进行替换,并将其后的所有值替换为X

若不存在这样的替换,则返回XX:XX

Python代码:

class Solution(object):
    def nextClosestTime(self, time):
        """
        :type time: str
        :rtype: str
        """
        time = time[:2] + time[3:]
        isValid = lambda t: int(t[:2]) < 24 and int(t[2:]) < 60
        stime = sorted(time)
        for x in (3, 2, 1, 0):
            for y in stime:
                if y <= time[x]: continue
                ntime = time[:x] + y + (stime[0] * (3 - x))
                if isValid(ntime): return ntime[:2] + ':' + ntime[2:]
        return stime[0] * 2 + ':' + stime[0] * 2

 

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