LeetCode-92-Reverse Linked List II 链表反转 各种边界条件

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def reverseBetween(self, head, m, n):        """        :type head: ListNode        :type m: int        :type n: int        :rtype: ListNode        """        if n==m:return head        totalHead=ListNode(0)        totalHead.next=head        left=totalHead        right=totalHead        for i in range(m-1):            left=left.next        tail=left.next        cur1=tail        cur2=tail.next        if cur2==None:return totalHead.next        cur3=tail.next.next        if cur3==None:            cur2.next=cur1            cur1.next=None            left.next=cur2            return totalHead.next        for i in range(n-m):            cur2.next=cur1            cur1=cur2            cur2=cur3            cur3=cur3.next            if cur3==None:                if i==n-m-1:break                cur2.next=cur1                left.next=cur2                tail.next=None                return totalHead.next                                        left.next=cur1        tail.next=cur2        return totalHead.next