# Leetcode Algorithms

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->NULLOutput: 1->3->5->2->4->NULL`

Example 2:

`Input: 2->1->3->5->6->4->7->NULLOutput: 2->3->6->7->1->5->4->NULL`

Constraints:

• 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 …
• The length of the linked list is between `[0, 10^4]`.

Logic:

search for the length of the linked list, and iterate it and then put the even at the end

Solution:

`class Solution:    def oddEvenList(self, head:ListNode) -> ListNode:    ```    :type head:ListNode    :rtype: ListNode    ```    if head is None:        return head    odd = oddHead = head    even = evenHead = head.next # put even at the end    while even and even.next:        odd.next = even.next        odd = odd.next        even.next = odd.next        even = even.next        odd.next = evenHead    return oddHead    `

Credit

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## Jen-Li Chen

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