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

題目鏈接

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

簡單的指針變換。因爲Python中沒有指針，所以一直沒有看過Python的“指針”。這部分的Python實現值得去深究一下。
創建兩個子鏈：奇鏈和偶鏈。逐一遍歷所有的節點，用一個mark位來標誌當前節點是奇還是偶，將節點接到奇鏈或偶鏈上。

AC代碼

# 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))

遺留問題

Python的“指針”是如何實現的。