Odd Even Linked List

Question

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: Given 1->2->3->4->5->NULL, return 1->3->5->2->4->NULL.

Note:

• 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 ...

Tags

• Linked List

Thought

Use two dummy nodes to split the linked list and concatenate them later.

Code

# 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
        """
        oddPtr = oddHead = ListNode(-1)
        evenPtr = evenHead = ListNode(-1)
        ptr = head
        while ptr:
            oddPtr.next = ptr
            oddPtr = oddPtr.next
            ptr = ptr.next
            if ptr:
                evenPtr.next = ptr
                evenPtr = evenPtr.next
                ptr = ptr.next
        oddPtr.next = evenHead.next
        evenPtr.next = None
        return oddHead.next