[LeetCode#201] Bitwise AND of Numbers Range

The idea behind this problem is not hard, you could easily think out the solution. But for the implementation, you may trap yourself into some sterotypes. 
Idea:
Since for bitwise ‘AND‘ operation, as long as there is a ‘0‘ appears, the digit should become 0.

Principle in digit change in range: (work for numerial system, binary system and other system too)
Start:  1011111100111010101010
End:    1011111110001010010101

To reach 10001010010101 from 00111010101010. We must start to add 
00000000000001 onto 00111010101010 => 00111010101011
...
We can we reach the ‘1‘ in high index, all digits behind after it must change once. That‘s to say, all the digits after 1 (include 1) should become 0.
answer: 1011111110000000000000

Algorithm:
Step 1: Find the first left bit ‘start‘ differ from ‘end‘.
Start:  10111111[0]0111010101010
End:    10111111[1]0001010010101

Step 2: Make all bits after the first different bits (include first different bit) into 0. The number is the answer we wish to get.
    10111111[0]0111010101010
=>  10111111000000000000

Great implementation:
Move bit to reach the answer!!!! See the solution!

public class Solution {
    public int rangeBitwiseAnd(int m, int n) {
        if (m < 0 || n < 0)
            throw new IllegalArgumentException("the passed in arguements is not tin the valid range!");
        int p = 0;
        while (m != n) {
            m = m >> 1;
            n = n >> 1;
            p++;
        }
        return m << p;
    }
}