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. 

    public int rangeBitwiseAnd(int m, int n) {
        if (m == 0) {
            return 0;
        }
        int count = 0;
        while (m != n) {
            m >>>= 1;
            n >>>= 1;
            count++;
        }
        return m <<= count;
    }