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:

  1. Elements of the given array are in the range of to 10^9
  2. Length of the array will not exceed 10^4.
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Using bit manipulation to achieve operators

1)计算负数的逆元

def negative(n):
    return bit_add(~n, 1)

2)加法计算

def bit_add(a, b):
    b &= (2 ** 32 - 1)
    while b:
        tmp = a
        sum = tmp ^ b
        carry = (tmp & b) << 1
    return a

3)减法计算

def bit_sub(a, b):
    return bit_add(a, negative(b))

4)乘法计算

def bit_mul(a, b):
    result = 0

    while b:
        if b & 0x1:
            result = bit_add(result, a)
        b >>= 1
        a <<= 1

    return result

乘法模拟手工计算乘法的流程,将乘数b的二进制表示从左往右移动,当最低位为1时 (b & 0x1 == 1),把被乘数a加到result中,同时将被乘数a向左移动一位。

5)除法计算

def bit_div(x, y):
    result = 0
    for i in range(32)[::-1]:
        if (x >> i) >= y:
            result += (1 << i)
            x -= (y << i)

    return result

考虑32位整数表示 (64位改一下 i 就可以了)。从最大的倍数开始遍历,如果被除数x >> i 大于除数y,说明该位商为1,将1<<i加到结果中,并缩小x。