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1 分鐘 閱讀文 (大約 112 個字)

Binary Number with Alternating Bits

題目

  1. 先想到的是有沒有辦法透過 bitwise operation 來判斷，發現無法
  2. 看相鄰位置有沒有一樣的，最直覺就是用字串來判斷 => 可以利用 Python 的 built-in function bin 來把數字轉為 binary 字串

Python3 solution:

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def hasAlternatingBits(self, n: int) -> bool:
    b_str = bin(n)[2:]  # 去掉前面的 '0b'
    for i in range(len(b_str) - 1):
        if b_str[i] == b_str[i + 1]:
            return False
    return True
#
Isomorphic Strings
Single Number

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android array authentication binary search binary search tree binary tree bit manipulation breadth-first search database depth-first search distributed system django docker dokku dynamic programming graph greedy hash table heap heroku java jwt linked list matrix mysql paas package management pipenv postgresql python recursion shortest path simulation sliding window sorting stack string tree two pointers

最新文章

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標籤

android 2
array 5
authentication 1
binary search 1
binary search tree 1
binary tree 6
bit manipulation 2
breadth-first search 5
database 2
depth-first search 9
distributed system 1
django 1
docker 3
dokku 1
dynamic programming 4
graph 2
greedy 1
hash table 6
heap 1
heroku 1
java 1
jwt 1
linked list 5
matrix 1
mysql 1
paas 1
package management 1
pipenv 1
postgresql 1
python 9
recursion 11
shortest path 1
simulation 1
sliding window 1
sorting 1
stack 2
string 4
tree 7
two pointers 3

最新文章

彙整

標籤

android 2
array 5
authentication 1
binary search 1
binary search tree 1
binary tree 6
bit manipulation 2
breadth-first search 5
database 2
depth-first search 9
distributed system 1
django 1
docker 3
dokku 1
dynamic programming 4
graph 2
greedy 1
hash table 6
heap 1
heroku 1
java 1
jwt 1
linked list 5
matrix 1
mysql 1
paas 1
package management 1
pipenv 1
postgresql 1
python 9
recursion 11
shortest path 1
simulation 1
sliding window 1
sorting 1
stack 2
string 4
tree 7
two pointers 3
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