Leetcode Bitwise and of Numbers Range

Given a range [M, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, Inclus Ive.

For example, given the range [5, 7], and you should return 4.

This is a bit operation.

One basic formula to explore is (N & (n-1))

Suppose there is such a number xxxxxxxxxxx10000 (this indicates that the rightmost one is a bit of 1, and X is 1 can be 0)

So n-1 = xxxxxxxxxxx01111

N & (n-1)?

We found that the 1 right side was eliminated.

Wait, there seems to be information to explore n = xxxxxxxxxxx10000, n-1 = xxxxxxxxxxx01111, N & (n-1) = xxxxxxxxxxx00000

Have you found that the difference between N and (N & (n-1)) is not more than 1 ah, so what is the number of this between the form??

n = xxxxxxxxxxx10000

N-1 = xxxxxxxxxxx01111

xxxxxxxxxxx01110

......

Xxxxxxxxxxx0xxxx

......

N & (n-1) = xxxxxxxxxxx00000

Did you find out that all these numbers and the result is N & (n-1)

So the solution is very concise:

    int Rangebitwiseand (int m, int n) {        while (n > M)            n &= n-1;        return n;    }

Leetcode Bitwise and of Numbers Range