[leetcode] 949. Largest Time for Given Digits

[leetcode] 949. Largest Time for Given Digits

Description

Given an array arr of 4 digits, find the latest 24-hour time that can be made using each digit exactly once.

24-hour times are formatted as “HH:MM”, where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour time is 00:00, and the latest is 23:59.

Return the latest 24-hour time in “HH:MM” format. If no valid time can be made, return an empty string.

Example 1:

Input: A = [1,2,3,4]Output: "23:41"Explanation: The valid 24-hour times are "12:34", "12:43", "13:24", "13:42", "14:23", "14:32", "21:34", "21:43", "23:14", and "23:41". Of these times, "23:41" is the latest.

Example 2:

Input: A = [5,5,5,5]Output: ""Explanation: There are no valid 24-hour times as "55:55" is not valid.

Example 3:

Input: A = [0,0,0,0]Output: "00:00"

Example 4:

Input: A = [0,0,1,0]Output: "10:00"

Constraints:

arr.length == 40 <= arr[i] <= 9

分析

题目的意思是：给定4个数，求能够组成的最大时间点的数。这道题我最先想到的是用递归，把所有的组合列举出来，选出合法的最大时间点就行了，遍历的时候用visited记录遍历过的点。这道题我看答案用的是暴力求解，调用了permutations函数，也可以。

代码

class Solution: def solve(self,arr,ans,visited): if(len(ans)==len(arr)): h=ans[0]*10+ans[1] s=ans[2]*10+ans[-1] if(h<24 and s<=59): if(self.max_hour str: n=len(arr) min_val=min(arr) self.max_hour=-1 self.max_sec=-1 visited=[False]*4 self.solve(arr,[],visited) if(self.max_hour==-1): return '' return "{:02d}:{:02d}".format(self.max_hour, self.max_sec)

参考文献

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