LeetCode刷题 (Python) | 328. Odd Even Linked List

题目链接

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:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->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 …

心得

简单的指针变换。因为Python中没有指针，所以一直没有看过Python的“指针”。这部分的Python实现值得去深究一下。
创建两个子链：奇链和偶链。逐一遍历所有的节点，用一个mark位来标志当前节点是奇还是偶，将节点接到奇链或偶链上。

AC代码

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


class Solution(object):
    def oddEvenList(self, head):
        """ :type head: ListNode :rtype: ListNode """
        if head is None or head.next is None:
            return head
        oddMark = True
        oddTail = head
        evenHead = head.next
        evenTail = head.next
        point = evenTail.next
        while point is not None:
            if oddMark:
                oddTail.next = point
                oddTail = point
            else:
                evenTail.next = point
                evenTail = point
            point = point.next
            oddMark = not oddMark
        evenTail.next = None
        oddTail.next = evenHead
        return head


def printLinkedList(head):
    while head != None:
        print(head.val, end=' ')
        head = head.next


def listToLinkedList(l):
    if len(l) == 0:
        return None
    head = ListNode(l[0])
    point = head
    for i in l[1::]:
        point.next = ListNode(i)
        point = point.next
    return head


if __name__ == '__main__':
    head = listToLinkedList([1, 2, 3, 4, 5, 6, 7, 8])
    s = Solution()
    printLinkedList(s.oddEvenList(head))

遗留问题

Python的“指针”是如何实现的。