标签:== bit org possible find with amp ast load
A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent the minutes (0-59).
Each LED represents a zero or one, with the least significant bit on the right.
For example, the above binary watch reads "3:25".
Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.
Example:
Input: n = 1
Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]
Note:
class Solution(object):
def readBinaryWatch(self, num):
"""
:type num: int
:rtype: List[str]
"""
ans=[]
def findans(leds,bits=0,i=0):
if bits==num:
h,m=int(‘‘.join(leds[:4]),2), int(‘‘.join(leds[4:]),2)
if h<12 and m<60:
ans.append("{}:{:02d}".format(h,m))
return
for j in range(i,10):
leds[j]=‘1‘
findans(leds,bits+1,j+1)
leds[j]=‘0‘
findans([‘0‘]*10)
return ans
[LeetCode&Python] Problem 401. Binary Watch
标签:== bit org possible find with amp ast load
原文地址:https://www.cnblogs.com/chiyeung/p/10109760.html