LeetCode 328. Odd Even Linked List

Posted on | In |
Words count in article: 243 | Reading time ≈ 1

题目

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:

1
2
Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL

Example 2:

1
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 …

思路

Put the odd nodes in a linked list and the even nodes in another. Then link the evenList to the tail of the oddList.

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head:
            return head
        odd = oddHead = head
        even = evenHead = head.next
        while even and even.next:
            odd.next = even.next
            odd = odd.next
            even.next = odd.next
            even = even.next
        odd.next = evenHead
        return oddHead