LeetCode-92. 反转链表 II【链表】
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
解题思路一：简单的翻转链表操作
python
# 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: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
p0 = dummy = ListNode(next=head)
for _ in range(left-1):
p0 = p0.next
pre = None
cur = p0.next
for _ in range(right - left + 1):
nxt = cur.next
cur.next = pre
pre = cur
cur = nxt
p0.next.next = cur # 让 转头 指向 后面
p0.next = pre # 让 前 指向 转尾
return dummy.next时间复杂度：O(n)
空间复杂度：O(1)
背诵版：
python
# 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: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
p0 = dummy = ListNode(next = head)
for _ in range(left - 1):
p0 = p0.next
pre = None
cur = p0.next
for _ in range(right - left + 1):
nxt = cur.next
cur.next = pre
pre = cur
cur = nxt
p0.next.next = cur
p0.next = pre
return dummy.next时间复杂度：O(n)
空间复杂度：O(1)
解题思路三：0
python
时间复杂度：O(n)
空间复杂度：O(n)
创作不易，观众老爷们请留步... 动起可爱的小手，点个赞再走呗 (๑◕ܫ￩๑) 
欢迎大家关注笔者，你的关注是我持续更博的最大动力
原创文章，转载告知，盗版必究
♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠ ⊕ ♠