Problem statement:

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL

Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL

Constraints:

• The relative order inside both the even and odd groups should remain as it was in the input.
• The first node is considered odd, the second node even and so on ...
• The length of the linked list is between [0, 10^4].

Solution

class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null) return null;
        var odd = head;
        var even = head.next;
        var 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;
    }
}

Time complexity:

O(n) There are total nn nodes and we visit each node once.

Space complexity:

O(1)All we need is the three pointers.

Approach: