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.

这道题我就不说什么了，偷瞄的别人的，这么绝的解法无论如何也想不出：https://leetcode.com/discuss/35057/share-my-simple-java-solution

大概思想：如果m！=n，那么range[m,n]所有数最后一位相与必定为0，因为[m,n]必定存在奇偶数，按照此方法通过将m,n同时右移一位即可得到倒数第二位的值，依次类推。

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