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

# 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：

# 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

时间复杂度：O(n)
空间复杂度：O(n)

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