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
源代码文件在这里.
