Leetcode 328 - Odd Even Linked List

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Algorithms, Leetcode, Python
Catalogue
  1. 1. Question
  2. 2. Analysis
  3. 3. Code

Question

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 …

Analysis

这个链表问题不是复杂，大致的思路就是将奇数序数的节点连在一起，将偶数序数的节点连在一起，最后将偶数链表接在奇数链表最后一个节点即可。相比于每次移动一个节点，使用奇偶两个 head 一次移动两个节点的遍历更加简洁，也不用多加判断语句和变量判断最后的节点序数是偶数还是奇数。

Code

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# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head:
            return None
        odd = head
        even = head.next
        evenHead = even
        while even and even.next:
            odd.next = even.next
            odd = odd.next
            even.next = odd.next
            even = even.next
        odd.next = evenHead
        return head
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