题目浅析

  • 想查看原题可以点击题目链接。

  • 简单地说，就是给一个链表，现在需要按照一定的规则重新排列。

思路分享

代码解答（强烈建议自行解答后再看）

  • 参考题解
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def get_mid(self, head) -> ListNode:
        left, right = head, head
        while right and right.next:
            left, right = left.next, right.next.next

        return left

    def reverse(self, head) -> ListNode:
        pre = None
        cur = head
        while cur:
            nxt = cur.next
            cur.next = pre
            pre = cur
            cur = nxt
        return pre

    def reorderList(self, head: Optional[ListNode]) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        mid = self.get_mid(head)
        mid = self.reverse(mid)
        left = head
        while mid.next:
            left_nxt = left.next
            mid_nxt = mid.next
            left.next = mid
            mid.next = left_nxt
            left = left_nxt
            mid = mid_nxt