题目描述
You are given the head pointer head of the singly linked list and two integers left and right, where left <= right. Please reverse the linked list nodes from position left to position right, and return the reversed linked list.
示例
输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]
思路分析
This question is actually very similar to the last question about "Reversing a Linked List". Last time, the entire linked list was reversed. This time, an interval is taken, and the part of the linked list within the interval is reversed. The idea is to first take out the linked list to be reversed from the original linked list, reverse it and then splicing it back to the original linked list. The idea is relatively simple, but the code is more cumbersome to implement, please see the code.
代码
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseBetween(self, head: ListNode, left: int, right: int) -> ListNode:
# 定义反转链表函数，此函数在“反转链表”那一题已经讲过
def reverse(node):
before = node
after = None
while before:
next = before.next
before.next = after
after = before
before = next
return after
# 接下来就是要找到要反转的链表部分
# 先定位反转链表部分的左节点
# 因为头节点有可能发生变化，使用虚拟头节点可以避免复杂的分类讨论
dummy_node = ListNode(-1)
dummy_node.next = head
pre = dummy_node
# 虚拟头节点向前移动 left - 1 步, 到达左节点的前一个结点
for i in range(left - 1):
pre = pre.next
# pre节点继续向前移动right - left + 1步，到达右节点
right_node = pre
for j in range(right - left + 1):
right_node = right_node.next
# 定位到左节点
left_node = pre.next
# 保留右节点右边的部分，用于后边反转之后的拼接
curr = right_node.next
# 切出反转链表部分
pre.next = None
right_node.next = None
# 反转链表
reversed_linklist = reverse(left_node)
# 将反转后的链表接入到原链表
pre.next = reversed_linklist
left_node.next = curr
return dummy_node.next
运行结果
