Binary gap

Given a positive integer N, find and return the longest distance between two consecutive 1's in the binary representation of N.

If there aren't two consecutive 1's, return 0.

 

Example 1:

Input: 22
Output: 2
Explanation: 
22 in binary is 0b10110.
In the binary representation of 22, there are three ones, and two consecutive pairs of 1's.
The first consecutive pair of 1's have distance 2.
The second consecutive pair of 1's have distance 1.
The answer is the largest of these two distances, which is 2.
Example 2:

Input: 5
Output: 2
Explanation: 
5 in binary is 0b101.
Example 3:

Input: 6
Output: 1
Explanation: 
6 in binary is 0b110.
Example 4:

Input: 8
Output: 0
Explanation: 
8 in binary is 0b1000.
There aren't any consecutive pairs of 1's in the binary representation of 8, so we return 0.
 

Note:

1 <= N <= 10^9

这题求某个数二进制格式下最长的两个相邻的1之间的距离。熟悉二进制的人很容易就能解出来。

class Solution {
public:
  int binaryGap(int N) {
      int gap = 0;
      int prev = -1;
      int index = 0;
      while (N != 0) {
          if ((N & 0x1) != 0) {
              // Already found 1
              if (prev != -1 && index - prev > gap) {
                  gap = index - prev;
              }

              prev = index;
          }
          ++index;
          N = N >> 1;
      }
      return gap;
  }
};