328. Odd Even Linked List

1. Description

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.
The first node is considered odd, and the second node is even, and so on.
Note that the relative order inside both the even and odd groups should remain as it was in the input.
You must solve the problem in O(1) extra space complexity and O(n) time complexity.

2. Example

Example 1

Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]

Example 2

Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

3. Constraints

• The number of nodes in the linked list is in the range [0, $10^4$].
• $-10^6$ <= Node.val <= $10^6$

4. Solutions

Double Pointers

n is the number of nodes in head
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    ListNode *oddEvenList(ListNode *head) {
        if (head == nullptr || head->next == nullptr) {
            return head;
        }

        ListNode *even_head = head->next;
        ListNode *odd = head, *even = head->next;
        while (odd->next != nullptr && even->next != nullptr) {
            odd->next = even->next;
            odd = odd->next;

            even->next = odd->next;
            even = even->next;
        }

        odd->next = even_head;
        return head;
    }
};