[Leetcode]@python 92. Reverse Linked List II
题目链接
https://leetcode.com/problems/reverse-linked-list-ii/
题目原文
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NUL
L, m = 2 and n = 4,
return 1->4->3->2->5->NULL
.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
题目大意
给定一个链表,翻转从位置m到位置n之间的元素。
解题思路
翻转链表的问题:1)记录翻转前面部分的链表 2)翻转指定位置的链表 3)拼接链表
代码
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseBetween(self, head, m, n):
"""
:type head: ListNode
:type m: int
:type n: int
:rtype: ListNode
"""
if not head or not head.next:
return head
tmp = head
before = None
for i in range(1, m):
before = tmp
tmp = tmp.next
# beforeM is the (m-1)th one in list before reverse, atM is the (m)th one.
posM = tmp
# reverse link from m to n
pre = tmp
tmp = tmp.next
for i in range(m + 1, n + 1):
t = tmp.next
tmp.next = pre
pre = tmp
tmp = t
posM.next = tmp
# end of the sublist
if before:
before.next = pre
# in case m == 1, i.e no one before m
else:
head = pre
return head