题解 | 链表内指定区间反转

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param m int整型 
# @param n int整型 
# @return ListNode类
#
class Solution:
    def reverseBetween(self , head: ListNode, m: int, n: int) -> ListNode:
        # write code here
        dummy = ListNode(0)
        dummy.next = head
        pre = dummy

        # 找到第 m-1 个节点
        for _ in range(m - 1):
            pre = pre.next

        # 反转区间
        curr = pre.next
        prev = None
        for _ in range(n - m + 1):
            next_node = curr.next
            curr.next = prev
            prev = curr
            curr = next_node

        # 连接链表
        pre.next.next = curr  # 将反转后的区间与 succ 连接
        pre.next = prev       # 将 pre 与反转后的区间连接

        return dummy.next