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-8
import time
import itertools

'''
一次判斷數組中所有元素的一位，從低到高。如果某位上，爲0的元素有m個，爲1的元素有n個，則該位會產生的Hamming distance爲m*n個。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,] #實驗表明打表與不打表 在leetcode上排名從擊敗34.66%到了63.21% (2017.3.19數據)
        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