LeetCode Total Hamming Distance

本文介绍了一种计算一组整数间所有可能配对的汉明距离之和的方法。通过逐位对比并利用位运算技巧,有效地解决了问题。文章提供了一个Python实现示例。

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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
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