"Leetcode" Bitwise and of Numbers Range (middle)

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.

Ideas:

First, we look for the same bits in the previous binary, and the different bits in the following phase must be 0.

For example: 1111001

1110011

From the middle of the 0011-1001 will go through 1000 so that four bits are 0

My Code:

intRangebitwiseand (intMintN) {intAns =0; intt = -;  while(T >=0&& ((M & (1<< t) = = (N & (1<<T))) {ans|= M & (1<<t); T--; }        returnans; }

Someone else's code: more streamlined

int rangebitwiseand (intint  n) {    int r=integer.max_value;      while ((m&r)! = (n&r))  r=r<<1;     return n&R;}

"Leetcode" Bitwise and of Numbers Range (middle)