LeetCode143. Reorder List

最新推荐文章于 2022-10-08 18:49:12 发布

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本文详细解析了如何将单链表 L: L0→L1→…→Ln-1→Ln 重排序为 L0→Ln→L1→Ln-1→L2→Ln-2→… 的算法实现。通过三个步骤：找到中点并断开链表、反转后半部分链表、交叉连接两部分链表，实现了链表的高效重排序。

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

这个题思路上不难,稍微复杂一点,分三步(以1->2->3->4为例):

1.找到中间的点,把两个链表断开 1->2 3->4

2.把后面的链表倒置 1->2 4->3

3.把两个链表交叉连接起来 1->4->2->3

代码如下:

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

class Solution(object):
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: None Do not return anything, modify head in-place instead.
        """
        slow = fast = head
        if not head or not head.next:return head
        
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        middle = slow.next
        slow.next = None
        
        prev=None
        while middle:
            cur = middle
            middle = middle.next
            cur.next = prev
            prev = cur
        
        rhead = head
        while head and prev:
            p=head.next
            q=prev.next
            head.next = prev
            prev.next = p
            
            head=p
            prev=q
            
        return rhead

需要注意的是while fast and fast.next:我之前想法是可以只写while fast.next:,但这样是不对的,因为fast如果为None会报错