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：
等价于求m与n二进制编码中同为1的前缀。

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

class Solution {
    public int rangeBitwiseAnd(int m, int n) {
        int ans = n;
        while (ans > m) {
            ans = ans & (ans - 1);
        }

        return ans;
    }
}