Leetcode 20191225
The Operators:
x « y
Returns x with the bits shifted to the left by y places (and new bits on the right-hand-side are zeros). This is the same as multiplying x by 2**y.
x » y
Returns x with the bits shifted to the right by y places. This is the same as //‘ing x by 2**y.
x & y
Does a “bitwise and”. Each bit of the output is 1 if the corresponding bit of x AND of y is 1, otherwise it’s 0.
x | y
Does a “bitwise or”. Each bit of the output is 0 if the corresponding bit of x AND of y is 0, otherwise it’s 1.
~ x
Returns the complement of x - the number you get by switching each 1 for a 0 and each 0 for a 1. This is the same as -x - 1.
x ^ y
Does a “bitwise exclusive or”. Each bit of the output is the same as the corresponding bit in x if that bit in y is 0, and it’s the complement of the bit in x if that bit in y is 1.
Just remember about that infinite series of 1 bits in a negative number, and these should all make sense.
Bitwise operators
191. Number of 1 Bits
https://leetcode.com/problems/number-of-1-bits/description/
class Solution:
def hammingWeight(self, n: int) -> int:
count, mask = 0, 0b1
for _ in range(32):
if n & mask != 0:
count += 1
mask <<= 1
return count
if __name__ == '__main__':
assert Solution().hammingWeight(0b00000000000000000000000000001011) == 3
32 ms 12.8 MB
136. Single Number
https://leetcode.com/problems/single-number/description/
class Solution:
def singleNumber(self, nums):
x = 0
for item in nums:
x = x ^ item
return x
if __name__ == '__main__':
assert Solution().singleNumber([4, 1, 2, 1, 2]) == 4
88 ms 15 MB
Alternative solutions:
class Solution:
def singleNumber(self, nums: List[int]) -> int:
import functools
return functools.reduce(lambda x, y: x ^ y, nums)
461. Hamming Distance
https://leetcode.com/problems/hamming-distance/
class Solution:
def hammingDistance(self, x, y):
raw, count, mask = x ^ y, 0, 1
for _ in range(32):
if raw & mask:
count += 1
mask <<= 1
return count
if __name__ == '__main__':
assert Solution().hammingDistance(1, 4) == 2
24 ms 12.8 MB