Description
https://leetcode.com/problems/odd-even-linked-list/
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 1:
Input:1->2->3->4->5->NULL
Output:1->3->5->2->4->NULL
Example 2:
Input: 2->1->3->5->6->4->7->NULL
Output:2->3->6->7->1->5->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 …
Explanation
Create an odd list and an even list, then combine together.
Python Solution
# 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:
dummy = ListNode(0)
odd_dummy = ListNode(0)
even_dummy = ListNode(0)
dummy.next = head
count = 1
current = head
odd = odd_dummy
even = even_dummy
while current != None:
if count % 2 == 0:
if count == 2:
even_dummy.next = current
even.next = current
even = even.next
else:
if count == 1:
odd_dummy.next = current
odd.next = current
odd = odd.next
current = current.next
count += 1
even.next = None
odd.next = even_dummy.next
dummy.next = odd_dummy.next
return dummy.next
- Time complexity: O(N).
- Space complexity: O(1).
One Thought to “LeetCode 328. Odd Even Linked List”