Class Solution
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:
1 <= n <= 109
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Constructor Summary
Constructors -
Method Summary
Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
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Constructor Details
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Solution
public Solution()
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Method Details
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binaryGap
public int binaryGap(int n)
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