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?
Python 解答:
1.修改值
# 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:
lis = []
i = 1
p = head
while i < left:
p = p.next
i += 1
q = p
while i <= right:
lis.append(q.val)
q = q.next
i += 1
while i > left:
p.val = lis[i-left-1]
p = p.next
i -= 1
return head
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