LeetCode Total Hamming Distance

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

Now your job is to find the total Hamming distance between all pairs of the given numbers.

Example:
Input: 4, 14, 2

Output: 6

Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case). So the answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
Note:
Elements of the given array are in the range of 0 to 10^9
Length of the array will not exceed 10^4.

# coding=utf-8import timeimport itertools'''一次判断数组中所有元素的一位，从低到高。如果某位上，为0的元素有m个，为1的元素有n个，则该位会产生的Hamming distance为m*n个。所有位的依次相加即可2147483647>10^9'''class Solution(object):    def totalHammingDistance(self, nums):        """        :type nums: list[int]        :rtype: int        """        table = [1,                 2,                 4,                 8,                 16,                 32,                 64,                 128,                 256,                 512,                 1024,                 2048,                 4096,                 8192,                 16384,                 32768,                 65536,                 131072,                 262144,                 524288,                 1048576,                 2097152,                 4194304,                 8388608,                 16777216,                 33554432,                 67108864,                 134217728,                 268435456,                 536870912,                 1073741824,                 2147483648,] #实验表明打表与不打表 在leetcode上排名从击败34.66%到了63.21% (2017.3.19数据)        s = 0        for j in xrange(0, 32):            m = n = 0            for i in nums:                if i&table[j] == 0:                    m += 1                else:                    n += 1            s += m * n        return s