LeetCode: Odd Even Linked List Solution

Approach

2 pointers: head for odd and head for even (in terms of original index)

Implementation

1var oddEvenList = function (head) {
2  if (!head || !head.next) {
3    return head
4  }
5
6  let oddHead = (oddTail = head)
7  let evenHead = (evenTail = head.next)
8
9  while (oddTail && oddTail.next && evenTail && evenTail.next) {
10    oddTail.next = oddTail.next.next
11    oddTail = oddTail.next
12
13    evenTail.next = evenTail.next.next
14    evenTail = evenTail.next
15  }
16
17  oddTail.next = evenHead
18
19  return oddHead
20}

References

Original problem

Similar problems

Split Linked List in Parts