LeetCode-5（反转列表I、II）

一、Reverse Linked List

反转链表

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    #递归      def reverseList(self, head):        """        :type head: ListNode        :rtype: ListNode        """        return self.helper(head,None)    def helper(self,node,prev):        if node:                        cur=node.next            node.next=prev            return self.helper(cur,node)        else:            return prev  #######################################################     #迭代     def reverseList(self, head):        """        :type head: ListNode        :rtype: ListNode        """        prev=None        while head:            curr=head            head=head.next            curr.next=prev            prev=curr        return prev         ###简洁        pre=None        curr=head        while curr:            curr.next,pre,curr=pre,curr,curr.next        return pre    

二、 Reverse Linked List II
给定链表和m,n两位置,反m-n位置之间的链表

class Solution(object):    def reverseBetween(self, head, m, n):        """        :type head: ListNode        :type m: int        :type n: int        :rtype: ListNode        """        if m==n:            return head        dummy=ListNode(0)        dummy.next=head        pre=dummy        for i in range(m-1):            pre=pre.next        curr=pre.next        rev=None        for i in range(n-m+1):               tmp=curr.next            curr.next=rev            rev=curr            curr=tmp     ##反转中间一段后连接           pre.next.next=curr        pre.next=rev                 return dummy.next

if m==n:            return head        dummy=ListNode(0)        dummy.next=head        pre=dummy        for i in range(m-1):            pre=pre.next        curr=pre.next        then=curr.next        ##不断把then换到pre后面        for i in range(n-m):                  curr.next=then.next            then.next=pre.next            pre.next=then            then=curr.next        return dummy.next