反转链表 II

问题

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

链接：https://leetcode-cn.com/problems/reverse-linked-list-ii

解答

que2.reverse() 反转部分列表内容

# 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:
        que1 = []
        while head:
            que1.append(head.val)
            head = head.next
        que2 = que1[left-1:right]
        que2.reverse()
        res = que1[:left-1]+que2+que1[right:]
        ping = cur = ListNode()
        for i in res:
            cur.next = ListNode(i)
            cur = cur.next
        
        return ping.next

标签：II,head,right,ListNode,que1,next,链表,CodingPark,left
来源： https://blog.csdn.net/weixin_38411989/article/details/117594200