leetcode_206. 反转链表

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-linked-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

#栈
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        ls=[]
        while(head):#压栈
            ls.append(head)
            head=head.next
        p1=p2=ListNode(0,None)
        while(ls):
            t=ls.pop()
            p2.next=t
            p2=p2.next
        p2.next=None    
        return p1.next

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        ls=[]
        if not head or not head.next :
            return head
        while(head):
            ls.append(head)
            head=head.next
        p=p1=ls.pop()
        while(ls):
            p.next=ls.pop()
            p=p.next
        p.next=None
        return p1

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        p=None#建立空节点
        while(head):
            t=head.next#当前节点的下一个
            head.next=p#当前节点的next
            p=head#后移
            head=t#后移
        return p

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        #递归
        if not head or not head.next:
            return head
        t=self.reverseList(head.next)
        head.next.next=head
        head.next=None
        return t