Class Solution
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public final class Solution2807 - Insert Greatest Common Divisors in Linked List\.
Medium
Given the head of a linked list
head, in which each node contains an integer value.Between every pair of adjacent nodes, insert a new node with a value equal to the greatest common divisor of them.
Return the linked list after insertion.
The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.
Example 1:
Input: head = 18,6,10,3
Output: 18,6,6,2,10,1,3
Explanation: The 1<sup>st</sup> diagram denotes the initial linked list and the 2<sup>nd</sup> diagram denotes the linked list after inserting the new nodes (nodes in blue are the inserted nodes).
We insert the greatest common divisor of 18 and 6 = 6 between the 1<sup>st</sup> and the 2<sup>nd</sup> nodes.
We insert the greatest common divisor of 6 and 10 = 2 between the 2<sup>nd</sup> and the 3<sup>rd</sup> nodes.
We insert the greatest common divisor of 10 and 3 = 1 between the 3<sup>rd</sup> and the 4<sup>th</sup> nodes.
There are no more adjacent nodes, so we return the linked list.
Example 2:
Input: head = 7
Output: 7
Explanation: The 1<sup>st</sup> diagram denotes the initial linked list and the 2<sup>nd</sup> diagram denotes the linked list after inserting the new nodes. There are no pairs of adjacent nodes, so we return the initial linked list.
Constraints:
The number of nodes in the list is in the range
[1, 5000].1 <= Node.val <= 1000
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final ListNodeinsertGreatestCommonDivisors(ListNode head)-
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Method Detail
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insertGreatestCommonDivisors
final ListNode insertGreatestCommonDivisors(ListNode head)
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