链表_leetcode92

#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
 """

 if not head or not head.next:
 return head
 if n < 1 or m >= n :
 return head

 cout = 1
 cur = head



 # 还是边界条件没考虑完全呀，比如翻转整个链表

 preStart = None
 nextEnd = None

 while cur:

 cur = cur.next

 cout += 1

 if cout == m-1:
 preStart = cur
 if cout == n:
 nextEnd = cur.next

 break


 start = preStart.next

 begin = start.next

 for i in range(m-n):

 nextBegin = begin.next
 begin.next = preStart.next
 preStart.next = begin
 begin = nextBegin

 start.next = nextEnd

 return head



class Solution2(object):
 def reverseBetween(self, head, m, n):
 """
 :type head: ListNode
 :type m: int
 :type n: int
 :rtype: ListNode
 """

 count = 1
 root = ListNode(0)
 root.next = head

 pre = root

 # while pre.next:
 #
 # pre = pre.next
 # count += 1
 #
 # if count == m :
 # break

 while pre.next and count < m :
 pre = pre.next
 count += 1

 if not pre:
 return head


 mNode = pre.next
 cur = mNode.next

 while cur and count < n:

 next = cur.next
 cur.next = pre.next
 pre.next = cur
 mNode.next = next

 cur = next
 count += 1



 return root.next