092 reverse linked list ii
92. Reverse Linked List II
题目: https://leetcode.com/problems/reverse-linked-list-ii/
难度: Medium
跟 reverse linked list一样
思路： 找到 第 m 个node，然后开始reverse到第n个node，然后再把它们和原本的list连接起来
AC 代码
class Solution(object):
def reverseBetween(self, head, m, n):
"""
:type head: ListNode
:type m: int
:type n: int
:rtype: ListNode
"""
# m == n, not reverse
if m == n : return head
dummy = ListNode(-1)
dummy.next = head
mbefore = dummy
cnt = 1
while mbefore and cnt < m:
mbefore = mbefore.next
cnt += 1
prev = None
cur = mbefore.next
tail1 = mbefore.next
while cnt <= n :
nxt = cur.next
cur.next = prev
prev = cur
cur = nxt
cnt += 1
mbefore.next = prev
tail1.next = cur
return dummy.next
看了一下别人的代码，又比我写的好嘛，因为是保证m和n有效，用的是for循环先找到 m node:
for _ in range(m-1):
....
for _ in range(n-m):
reverse 操作
