leetcode 461. Hamming Distance
2017-02-13 22:13
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leetcode 461. Hamming Distance
题目
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.
solution:
class Solution { public: int hammingDistance(int x, int y) { int z = x^y; int ans = 0; while(z){ ans++; z &= z - 1; } return ans; } };
写此博客主要是想记录z &= z - 1位运算的巧妙运用:
n = 10100(二进制),则(n-1) = 10011 ==》n&(n-1) = 10000
可以看到原本最低位为1的那位变为0。借此可以有以下应用:
1: 判断一个数是否是2的方幂(次数二进制中只有最高位为1)
n > 0 && ((n & (n - 1)) == 0 )
2:计算一个正整数的二进制中含有1的个数,例如本文代码中
3:计算N!的质因数2的个数
容易得出N!质因数2的个数 = [N / 2] + [N / 4] + [N / 8] + ….
下面通过一个简单的例子来推导一下过程:N = 10101(二进制表示)
现在我们跟踪最高位的1,不考虑其他位假定为0,
则在
[N / 2] 01000
[N / 4] 00100
[N / 8] 00010
[N / 8] 00001
则所有相加等于01111 = 10000 - 1
由此推及其他位可得:(10101)!的质因数2的个数为10000 - 1 + 00100 - 1 + 00001 - 1 = 10101 - 3(二进制表示中1的个数)
推及一般N!的质因数2的个数为N - (N二进制表示中1的个数)
[1]n & (n-1)
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