Bitwise AND of Numbers Range

题目：

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.

思路：

To get a better preparation for interviews, I decided to explain my ideas in English. 

Here is what I thought: 

1) For any given m and n, as long as m != n, we always get 0 at right-most bit when we do AND operation on these numbers. It is because two adjacent numbers always have different right-most bit. There always be a 0!

   like:  5: 0101

　　　　6: 0110

　result:   0100

2) So, we keep doing right shift until m is equal to n. And we count the number of times of right shift(keep this number in "count").  

3) We do left shift to m. Do it "count" times. 

 1     public int rangeBitwiseAnd(int m, int n) {
 2         if (m == 0) {
 3             return 0;
 4         }
 5         int count = 0;
 6         while (m != n) {
 7             m >>>= 1;
 8             n >>>= 1;
 9             count++;
10         }
11         return m <<= count;
12     }

Bitwise AND of Numbers Range

原文：http://www.cnblogs.com/gonuts/p/4436574.html