LeetCode||92. Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:

Given 

1->2->3->4->5->NULL

, m = 2 and n = 4,

return 

1->4->3->2->5->NULL

.

Note:

Given m, n satisfy the following condition:

1 ≤ m ≤ n ≤ length of list.
交换链表上的元素
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None

class Solution(object):
def reverseBetween(self, head, m, n):
"""
:type head: ListNode
:type m: int
:type n: int
:rtype: ListNode
"""
dummy = ListNode(0)
dummy.next = head
cur = dummy
for i in range(1, m):
cur = cur.next

tail = cur.next
for i in range(m, n):
if tail is None:
break
tmp = tail.next
tail.next = tmp.next
tmp.next = cur.next
cur.next = tmp

return dummy.next