n & (n - 1)的作用：把n中为1的最低位变为0.
Power of Two
Given an integer, write a function to determine if it is a power of two.
解法：2的次幂的特征是有且仅有一位为1，其余为均为零。故可利用判断n & (n - 1)是否为零来判断n是否为2的次幂。
public class Solution { public boolean isPowerOfTwo(int n) { return (n > 0) && ((n & (n - 1)) == 0); } }
Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1‘ bits it has (also known as the Hamming weight).
For example, the 32-bit integer ’11‘ has binary representation 00000000000000000000000000001011, so the function should return 3.
解法：根据n & (n - 1)的执行次数判断1的个数，直到为0.
public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int result = 0; while (n != 0) { result++; n &= (n - 1); } return result; } }
还有一种解法：
public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int result = 0; while (n != 0) { result += n & 1; n >>>= 1; } return result; } }
Bitwise AND of Numbers Range
Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.
For example, given the range [5, 7], you should return 4.
本题的目标是把m和n从最高位开始连续相同的位数不变，第一位不相同的位数开始都变为0.
解法一：找出低位中的1，将1置0.
public class Solution { public int rangeBitwiseAnd(int m, int n) { while (n > m) { n &= (n - 1); } return n; } }
解法二：迭代依次将低位置0.
public class Solution { public int rangeBitwiseAnd(int m, int n) { if (m == n) { return m; } if (m == n - 1) { return m & n; } return rangeBitwiseAnd(m / 2, n / 2) << 1; } }
原文：http://www.cnblogs.com/shinning/p/4991191.html