题目地址(92. 反转链表 II)
https://leetcode-cn.com/problems/reverse-linked-list-ii/
题目描述
给你单链表的头指针 head 和两个整数 left 和 right ,其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点,返回 反转后的链表 。
示例 1:
输入:head = [1,2,3,4,5], left = 2, right = 4
输出:[1,4,3,2,5]
示例 2:
输入:head = [5], left = 1, right = 1
输出:[5]
提示:
链表中节点数目为 n
1 <= n <= 500
-500 <= Node.val <= 500
1 <= left <= right <= n
进阶: 你可以使用一趟扫描完成反转吗?
前置知识
公司
- 暂无
思路
关键点
代码
解法一:
- 语言支持:Python3
Python3 Code:
# 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:
def reverse_linked_list(head:ListNode):
pre = None
cur = head
while cur:
next = cur.next
cur.next = pre
pre = cur
cur = next
# 因为头节点可能发生变化 因此我们采用虚拟节点可以避免复杂的分类讨论
dummy_node = ListNode(-1)
dummy_node.next = head
pre = dummy_node
# 找到反转区间的pre 走left-1步
for _ in range(left-1):
pre = pre.next
# 从pre再走right - left + 1步,来到right节点
right_node = pre
for _ in range(right-left+1):
right_node = right_node.next
# 取出子链表
left_node = pre.next
curr = right_node.next
# 切断连接
pre.next = None
right_node.next = None
# 反转子链表
reverse_linked_list(left_node)
# 接回到原链表
pre.next = right_node
left_node.next = curr
return dummy_node.next
复杂度分析
令 n 为数组长度。
- 时间复杂度: O ( n ) O(n) O(n)
- 空间复杂度: O ( 1 ) O(1) O(1)
解法二:
Python3 Code:
# 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:
# 由于头节点的不确定性 因此引入虚拟节点
dummy_node = ListNode(-1)
# 找到pre cur
dummy_node.next = head
pre = dummy_node
for _ in range(left-1):
pre = pre.next
cur = pre.next
for _ in range(right-left):
next = cur.next
cur.next = next.next
next.next = pre.next
pre.next = next
return dummy_node.next
复杂度分析
令 n 为数组长度。
- 时间复杂度: O ( n ) O(n) O(n)
- 空间复杂度: O ( 1 ) O(1) O(1)