leetcode 461. Hamming Distance

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 231.

Example:

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.

解法1：

class Solution(object):
    def hammingDistance(self, x, y):
        """
        :type x: int
        :type y: int
        :rtype: int
        out(0, 0)=0
        out(0, 1)=1
        out(1, 0)=1
        out(1, 1)=0
        out(2, 1)=2
        out(4, 1)=2
        out(0xffffffff, 0)=31?
        1100 & 1011 = 1000
        """
        z = x ^ y
        res = 0
        while z:            
            res += 1
            z = z & (z-1)
        return res
        

解法2：

    def hammingDistance(self, x, y):
        """
        :type x: int
        :type y: int
        :rtype: int
        """
        return bin(x^y).count('1')

因为：

>>> bin(12)
'0b1100'
>>> '0b1100'.count('1')
2
>>> 'abcb'.count('b')
2

解法3：

We can find the i-th bit (from the right) of a number by dividing by 2 i times, then taking the number mod 2.

Using this, lets compare each of the i-th bits, adding 1 to our answer when they are different.

Code:

ans = 0
while x or y:
  ans += (x % 2) ^ (y % 2)
  x /= 2
  y /= 2
return ans

比较笨拙，其实 (x & 1) ^ (y & 1), x>>1，y>>1 更好！单独判断某位是否为1移位即可，当然用%2也可以。

        res = 0
        while x or y:
            res += (x&1) ^ (y&1)
            x = x>>1
            y = y>>1
        return res