LeetCode Problem 92-Reverse Linked List II

反转链表 II。反转从位置 mn 的链表。请使用一趟扫描完成反转。

说明: 1 ≤ mn ≤ 链表长度。

示例:

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输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL
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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

思路一

首先定位第 个节点,然后完成下面的操作即可:

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Step 1: 1->2->3->4->5
           p     q
Step 2: 1->3->4->2->5  将结点p移动到结点q后面
           p  q
Step 3: 1->4->3->2->5  将结点p移动到结点q后面

时间复杂度

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class Solution:
    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
        dummy = ListNode(-1)
        dummy.next = head
        p = dummy
        for i in range(m - 1):
            p = p.next
        tail = head = p.next
        for i in range(n - m):
            tail = tail.next
        for i in range(n - m):
            q = head.next
            head.next = tail.next
            tail.next = head
            head = q
        p.next = head
        return dummy.next

思路二

首先定位第 个节点,然后完成下面的操作即可:

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Step 1: 1->2->3->4->5
        p  c  q
Step 2: 1->3->2->4->5  将结点q移动到结点p后面
        p     c  q
Step 3: 1->4->3->2->5  将结点q插入到结点p后面

这里与思路一的区别是不用找到第 个结点,省去了一次循环,时间复杂度

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class Solution:
    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
        dummy = ListNode(-1)
        dummy.next = head
        pre = dummy
        for i in range(m - 1):
            pre = pre.next
        cur = pre.next
        after = cur.next
        for i in range(n - m):
            cur.next = after.next
            after.next = pre.next
            pre.next = after
            after = cur.next
        return dummy.next

相似问题

  1. Reverse Linked List
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