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.
Input: 1->2->3->4->5->NULL Output: 1->3->5->2->4->NULL
Input: 2->1->3->5->6->4->7->NULL Output: 2->3->6->7->1->5->4->NULL
- 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 ...
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def oddEvenList(self, head: ListNode) -> ListNode:
if not head:
return None
even_head = head.next
curr = head
is_odd = True
while curr.next:
tmp = curr.next
curr.next = tmp.next
prev = curr
curr = tmp
is_odd = not is_odd
if is_odd:
curr.next = even_head
else:
prev.next = even_head
return head