class Solution(object):
def isPowerOfTwo(self, number):
"""
input: int number
return: boolean
"""
# write your solution here
if number<=0:
return False
while (number&1 ==0):
number = self.rshift(number,1)
return number == 1
def rshift(self, val, n):
return (val & 0X7FFFFFFF) >> n
Determine the # of bits are different between 2 positive integers
xor 然后取多少个1
class Solution(object):
def diffBits(self, a, b):
"""
input: int a, int b
return: int
"""
# write your solution here
a ^= b
count = 0
while a != 0:
count += a & 1
a = self.rshift(a, 1)
return count
def rshift(self, val, n):
return (val & 0X7FFFFFFF) >> n
如何数1: Hamming Weight
What happens if we assign a negative number to an unsigned integer
class Solution(object):
def allUnique(self, word):
"""
input: string word
return: boolean
"""
# write your solution here
bitmap = [0]*8
for char in word:
idx = ord(char)
row = idx//32
col = idx%32
bitpos = (1<<col)
if bitmap[row]&bitpos!=0:
return False
else:
bitmap[row]|=bitpos
return True
int[] bitMap = new int[8];
for (int i=0; i<a.length();i++) {
int idx=a.charAt(i); //不需要减‘a’ 因为我们是针对整个ASCII表里本身的真实位置
int row = idx/32;
int col = idx%32;
int bitPos = (1 << col); //x&(1<<k) or (x>>k)&1
if (bitMap[row] & bitPos != 0) {
return false;
} else {
bitMap[row] |= bitPos;
}
}
return true;
How to reverse all bits of a number
用2 pointers，reverse string的方法
记得swap后赋值，因为int是primitive type
swap里的操作，交换，如果
XOR 1就是取反
class Solution(object):
def reverseBits(self, n):
"""
input: long n
return: long
"""
# write your solution here
for shift in range (16):
right = (n >> shift) &1
left = (n >> (31-shift)) &1
if left!=right:
n ^= (1 <<shift)
n ^= (1 <<(31-shift))
return n
10进制Number to 16进制
其实是0b到0X因为计算机里都是2进制，所以可以每四组一起，8421一个个抠出来；
怎么扣？(>>28)&0xF (>>24)&0xF
#这个有点问题
class Solution(object):
def hex(self, number):
"""
input: int number
return: string
"""
# write your solution here
result = []
if number==0:
return '0x0'
while number>0:
res = number%16
if res<10:
result.append(chr(res))
else:
result.append(chr(res-10+ord('A')))
number = self.rshift(number, 4)
result.append('x')
result.append('0')
return ''.join(result[::-1])
def rshift(self, val, n):
return (val & 0X7FFFFFFF) >> n
