leetcode 92. 反转链表 II 区间反转（不是整条链表反转）

1. 找到区间的pre、left、right、nxt，保存
2. 区间反转
3. pre.next 连接新头（right）
4. 区间尾巴（原来的left）接上nxt

# 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]:

        # 1. 找区间的pre、next 节点

        pre = ListNode(0)
        pre.next = head
        dummy = pre

        for i in range(left-1):
            pre = pre.next
        # 此时pre.next = 区间头节点
        left_node = pre.next

        right_node = head
        for i in range(right-1):
            right_node = right_node.next

        # 2. 反转区间
        cur = left_node

        # 断首尾
        prev = ListNode(0)
        prev.next = left_node   # 给个暂时的节点，指向区间left
        nxt = right_node.next    # 保存原始区间next

        while cur != nxt:  # 结束时cur已经出区间了，这样区间内才反转完
            tmp = cur.next  # 保存原始区间next
            cur.next = prev
            prev = cur
            cur = tmp

        pre.next = right_node  # 原来的区间pre_node 指向原来的尾节点（现在的left）
        left_node.next = nxt  # 现在的尾节点（原来的left）指向保存好的nxt

        return dummy.next