LeetCode 92. Reverse Linked List II.jpg
LeetCode 92. Reverse Linked List II
Description
Reverse a linked list from position m to n. Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.
Example:
Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL
描述
- 给定一个链表，反转指定的子序列.
# -*- coding: utf-8 -*-
# @Author: 何睿
# @Create Date: 2018-12-26 11:44:18
# @Last Modified by: 何睿
# @Last Modified time: 2018-12-26 13:36:47
class ListNode():
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def reverseBetween(self, head, m, n):
"""
:type head: ListNode
:type m: int
:type n: int
:rtype: ListNode
"""
# 需要交换的次数
move, reverse = m-1, n-m
# pre初始化指向第一个被换节点的前一个节点，point初始化为第一个被换节点
# 声明一个辅助头节点
res = ListNode(0)
pre, point = res, head
pre.next = head
# 把pre，point放到指定位置
while move > 0:
pre = pre.next
point = point.next
move -= 1
# 需要交换的次数
while reverse:
temp = point.next
point.next = temp.next
temp.next = pre.next
pre.next = temp
reverse -= 1
return res.next
