Given the head of a singly linked list, group all odd-indexed nodes together followed by the even-indexed nodes. The first node is odd (index 1), second even (index 2), etc.

Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]
Explanation: Odd nodes: 1→3→5, even nodes: 2→4. Join them. Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

Use two pointers: odd and even. Link odd.next = even.next, advance odd; link even.next = odd.next, advance even. Finally connect odd tail to even head.

public ListNode oddEvenList(ListNode head) { if (head == null) return null; ListNode odd = head, even = head.next, evenHead = even; while (even != null && even.next != null) { odd.next = even.next; odd = odd.next; even.next = odd.next; even = even.next; } odd.next = evenHead; return head; }
Complexity Analysis:

Time complexity: O(n) — single pass
Space complexity: O(1) — in-place, no extra space