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 …
1. Two head
# Time: O(n) # Space: O(1) # 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 or not head.next: return head odd, even = ListNode(0), ListNode(0) odd_head, even_head = odd, even idx = 0 while head: if idx & 1 == 0: odd.next = head odd = odd.next else: even.next = head even = even.next head = head.next idx += 1 even.next = None odd.next = even_head.next return odd_head.next # 71/71 cases passed (44 ms) # Your runtime beats 51.11 % of python3 submissions # Your memory usage beats 100 % of python3 submissions (14.6 MB)