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.
Input: head = [1,2,3,4,5] Output: [1,3,5,2,4]
Input: head = [2,1,3,5,6,4,7] Output: [2,3,6,7,1,5,4]
- The number of nodes in the linked list is in the range
-106 <= Node.val <= 106
Follow up: Could you solve it in
O(1) space complexity and
O(nodes) time complexity?
Create an odd list and an even list, then combine together.
# 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: odd_dummy = ListNode(0) even_dummy = ListNode(0) odd_node = odd_dummy even_node = even_dummy index = 0 while head != None: if index % 2 == 0: odd_node.next = ListNode(head.val) odd_node = odd_node.next else: even_node.next = ListNode(head.val) even_node = even_node.next head = head.next index += 1 odd_node.next = even_dummy.next return odd_dummy.next
- Time complexity: O(N).
- Space complexity: O(1).