Question

Formatted question description: https://leetcode.ca/all/328.html

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:

The number of nodes in the linked list is in the range [0, 10⁴].
-10⁶ <= Node.val <= 10⁶

Algorithm

Two parity pointers are used to point to the starting position of the even node respectively, and a separate pointer even_head is needed to save the starting position of the even node.

Then point the odd node to the next one of the even node (it must be the odd node), and move this odd node one step later.

Then point the even node to the next odd node (it must be an even node), this even node is moved one step later, and so on until the end.

At this time, the linked list of the separated even node is connected after the linked list of the odd node.

Code

public class Odd_Even_Linked_List {
    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */

    class Solution {
        public ListNode oddEvenList(ListNode head) {

            if (head == null) {
                return head;
            }

            // input head is odd head
            ListNode evenHeadDummy = new ListNode(0);

            ListNode oddPrev = head;
            ListNode evenPrev = evenHeadDummy;
            ListNode current = head.next;
            boolean isEven = true; // start from 2nd

            while (current != null) {

                if (isEven) {
                    evenPrev.next = current;
                    evenPrev = current;
                } else {
                    oddPrev.next = current;
                    oddPrev = current;
                }

                // flip
                isEven = !isEven;
                current = current.next;
            }

            // @note: cut last even->odd, or else loop
            evenPrev.next = null;

            // connect
            oddPrev.next = evenHeadDummy.next;

            return head;
        }
    }
}

############

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null) {
            return head;
        }
        ListNode odd = head, even = head.next;
        ListNode 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;
    }
}

// OJ: https://leetcode.com/problems/odd-even-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        ListNode evenHead, *evenTail = &evenHead, oddHead, *oddTail = &oddHead;
        for (bool odd = true; head; odd = !odd, head = head->next) {
            if (odd) {
                oddTail->next = head;
                oddTail = head;
            } else {
                evenTail->next = head;
                evenTail = head;
            }
        }
        oddTail->next = evenHead.next;
        evenTail->next = NULL;
        return oddHead.next;
    }
};

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        if head is None:
            return head
        odd, even = head, head.next
        even_head = even
        while even and even.next:
            odd.next = even.next
            odd = odd.next
            even.next = odd.next
            even = even.next
        odd.next = even_head
        return head

############

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
  def oddEvenList(self, head):
    """
    :type head: ListNode
    :rtype: ListNode
    """
    o = odd = ListNode(-1)
    e = even = ListNode(-1)
    p = head
    isOdd = True
    while p:
      if isOdd:
        o.next = p
        o = o.next
        isOdd = False
      else:
        e.next = p
        isOdd = True
        e = e.next
      p = p.next
    e.next = None
    o.next = even.next
    return odd.next

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func oddEvenList(head *ListNode) *ListNode {
	if head == nil {
		return head
	}
	odd, even := head, head.Next
	evenHead := even
	for even != nil && even.Next != nil {
		odd.Next = even.Next
		odd = odd.Next
		even.Next = odd.Next
		even = even.Next
	}
	odd.Next = evenHead
	return head
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function oddEvenList(head: ListNode | null): ListNode | null {
    if (head == null) return head;
    let odd: ListNode = head,
        even: ListNode = head.next;
    let 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;
}

328 - Odd Even Linked List

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