0328. Odd Even Linked List

https://leetcode.com/problems/odd-even-linked-list

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.

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]

Constraints:

• n ==number of nodes in the linked list

• 0 <= n <= 104

• -106 <= Node.val <= 106

ac

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        // edge cases
        if (head == null || head.next == null) return head;

        // pointers
        ListNode oh = head, ot = head, eh = head.next, et = head.next, r = head.next.next;
        et.next = null; // break list
        boolean odd = true;

        while (r != null) {
            ListNode tmp = r;
            r = r.next;

            if (odd) {  // if odd
                ot.next = tmp;
                tmp.next = eh;
                ot = ot.next;
            } else { // even
                et.next = tmp;
                tmp.next = null;
                et = et.next;
            }

            odd = !odd;
        }

        return head;
    }
}

/*
even head, even tail
odd head, odd tail
ranger
odd = true / false;
*/