题目
Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
解题思路
先把m和n转换成2进制，如果长度不同，则结果为0
找出m和n的公共前缀，后面变成0，就是最后的结果
代码
class Solution:
def rangeBitwiseAnd(self, m: int, n: int) -> int:
if len(bin(m)) != len(bin(n)):
return 0
res = 0
while (m != n):
m >>= 1
n >>= 1
res += 1
return m << res
