Reverse a linked list from position m to n. Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.
添加空头节点，可以减少好多边界判断。然后利用双指针，指针 p 指向反转段的上一个节点，指针 q 指向反转段的最后一个节点。
遍历原链表，如果遍历的长度 l < m，则 p 指向该节点；如果 l == m，则 p 保持不变不往后移动了，q 指向当前节点； 如果 m < l <= n，把该节点插入到 p 节点后边；如果 l > n，则把剩下的直接加到 q 的后边。
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
dummy = ListNode(-1)
p = dummy
cur = head
for i in range(m - 1):
p.next = cur
p = p.next
cur = cur.next
q = cur
for i in range(n - m + 1):
next_ = cur.next
cur.next = p.next
p.next = cur
cur = next_
q.next = cur
return dummy.next