201. Bitwise AND of Numbers Range

I brute force the problem in first place but I got TLE. Then I had to look at the problem closely. Look at 7,8,9,10,11,12,13,14 which have binary representation 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110. You'll find that the smallest number cancels the largest number's more significant bits.  And also, the adjacent number cancels the least significant bits which means, as long as a number is larger than the other, the least significant bit will get cancelled. So we can keep moving the least significant bits out and recall the function itself recursively until the two numbers are equal. The rest bits that the two equal number shares will be the result.

1:  class Solution {  
2:  public:  
3:    int rangeBitwiseAnd(int m, int n) {  
4:      return n > m ? (rangeBitwiseAnd(m>>1, n>>1)<<1) : m;  
5:    }  
6:  };