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Medium
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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 andO(n)
time complexity.
# 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: Optional[ListNode]) -> Optional[ListNode]:
if not head:
return head
odd=head
even=head.next
ehead=even
while even and even.next:
odd.next=odd.next.next
even.next=even.next.next
odd=odd.next
even=even.next
odd.next=ehead
return head