leetcode 143. Reorder List（python）

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描述

You are given the head of a singly linked-list. The list can be represented as:

L₀ → L₁ → … → L_{n - 1} → L_n

Reorder the list to be on the following form:

L₀ → L_n → L₁ → L_{n - 1} → L₂ → L_{n - 2} → …

You may not modify the values in the list's nodes. Only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4]
Output: [1,4,2,3]

Example 2:

Input: head = [1,2,3,4,5]
Output: [1,5,2,4,3]

Note:

The number of nodes in the list is in the range [1, 5 * 10^4].
1 <= Node.val <= 1000

解析

根据题意，给定了一个单向链表的头指针 head 。 该列表可以表示为：

L₀ → L₁ → … → L_{n - 1} → L_n

题目要求我们将链表重新排序为以下形式，不能修改节点中的值，只能更改节点本身：

L₀ → L_n → L₁ → L_{n - 1} → L₂ → L_{n - 2} → …

其实看完题意，思路还是比较容易想出来的，将原链表以中间的节点为中心，一分为二，然后将第二个链表进行反转，然后同时从左往右遍历两个链表，每次从第一个链表取一个节点拼接到新的链表，再从第二个链表取一个节点拼接到新的链表。最后得到的链表就是答案。考察的都是链表的基本操作，比如找中间节点，反转链表，拼接链表，遍历链表等等。

解答

class ListNode(object):
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution(object):
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: None Do not return anything, modify head in-place instead.
        """
        if not head or not head.next:return head
        node = self.getMid(head)
        head2 = node.next
        node.next = None
        head2 = self.getReverse(head2)
        self.merge(head, head2)
    
    def getMid(self, head):
        fast = head
        slow = head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        return slow
    
    def getReverse(self, head):
        cur = head
        last = None
        while cur:
            nxt = cur.next
            cur.next = last
            last = cur
            cur = nxt
        return last
    
    def merge(self, head1, head2):
        result = ListNode(0)
        while head1 or head2:
            if head1:
                result.next = head1
                head1 = head1.next
                result = result.next
            if head2:
                result.next = head2
                head2 = head2.next
                result = result.next

        	      
		

运行结果

Runtime: 88 ms, faster than 90.35% of Python online submissions for Reorder List.
Memory Usage: 31.3 MB, less than 48.45% of Python online submissions for Reorder List.

原题链接：leetcode.com/problems/re…

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