Solution

All Implemented Interfaces:
```
public final class Solution

                    
```
868 - Binary Gap.
Easy
Given a positive integer n, find and return the longest distance between any two adjacent 1's in the binary representation of n. If there are no two adjacent 1's, return 0.
Two 1's are adjacent if there are only 0's separating them (possibly no 0's). The distance between two 1's is the absolute difference between their bit positions. For example, the two 1's in "1001" have a distance of 3.
Example 1:
Input: n = 22
Output: 2
Explanation: 22 in binary is "10110".
The first adjacent pair of 1's is "10110" with a distance of 2.
The second adjacent pair of 1's is "10110" with a distance of 1.
The answer is the largest of these two distances, which is 2.
Note that "10110" is not a valid pair since there is a 1 separating the two 1's underlined.
Example 2:
Input: n = 8
Output: 0
Explanation: 8 in binary is "1000". There are not any adjacent pairs of 1's in the binary representation of 8, so we return 0.
Example 3:
Input: n = 5
Output: 2
Explanation: 5 in binary is "101".
Constraints:
- <code>1 <= n <= 10<sup>9</sup></code>

Nested Class Summary

Nested Classes
Modifier and Type	Class	Description

Field Summary

Fields
Modifier and Type	Field	Description

Constructor Summary

Constructors
Constructor	Description
`Solution()`

Enum Constant Summary

Enum Constants
Enum Constant	Description

Method Summary

Modifier and Type	Method	Description
`final Integer`	`binaryGap(Integer n)`

Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Detail
- Solution
```
Solution()
```

Method Detail
- binaryGap
```
 final Integer binaryGap(Integer n)
```