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 != 0) {

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

            break;

        }

        mask <<= 1;

    }



    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.