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;

    }

}