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.
Suppose our given range is 5 : 0101 ---- 7: 0111
Thus, we can see that the lower 2 bits are different. By shifting them to right, we can finally make m == n. thus, result will be m << count;
Code copied from the discussion. But this method is very fantastic!!
int rangeBitwiseAnd(int m, int n) { int count = 0; while(m != n) //until the left identical bits, all the right will be cancelled by increments from m to n;; { m >>= 1; n >>= 1; count++; } return m << count; }