201. Bitwise AND of Numbers Range -- 连续整数按位与的和

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.

int rangeBitwiseAnd(int m, int n) {

    int mask = 0xffffffff;

    /* find out the same bits in left side*/

    while (mask != ) {

        if ((m & mask) == (n & mask)) {

            break;

        }

        mask <<= ;

    }

    return m & mask;

}

Idea:

1) we know when a number add one, some of the right bit changes from 0 to 1 or  from 1 to 0

2) if a bit is 0, then AND will cause this bit to 0 eventually.

So, we can just simply check how many left bits are same for m and n.

for example:

5 is 101

6 is 110

when 5 adds 1, then the right two bits are changed.  the result is 100

6 is 110

7 is 111

when 6 adds 1, then the right one bit is changed. the result is 110.

9 is 1001

10 is 1010

11 is 1011

12 is 1100

Comparing from 9 to 12, we can see the first left bit is same, that's result.