单链表反转【python实现】
单链表反转方法
头节点开始遍历
- 用三个指针来解决链表反转,
cur
是现在的链表节点,tmp
暂存cur.next
,new_head
指向之前的cur
- 其中,
tmp
是保存之前的链条不能断的,很重要
尾节点开始递归
- 递归反转链表,递归:从尾节点指向上一级节点,上一级节点指向None,重复这个过程,直到递归结束
代码如下:
class ListNode:
def __init__(self, val):
self.val = val
self.next = None
class Solution:
def reverse_list(self, head: ListNode) -> ListNode:
if head == None or head.next == None:
return head
# 用三个指针来解决链表反转,cur是现在的链表节点,tmp暂存cur.next,new_head指向之前的cur
cur = head
tmp = None
new_head = None
while cur:
tmp = cur.next
cur.next = new_head
new_head = cur
cur = tmp
return new_head
def reverse_list2(self, head: ListNode) -> ListNode:
if head == None or head.next == None:
return head
# 通过Node指向之前的head来进行链表反转
Node = None
while head:
p = head
head = head.next
p.next = Node
Node = p
return Node
def reverse_list_recusive(self, head: ListNode) -> ListNode:
if not head or not head.next:
return head
# 递归反转链表,递归:从下一级的节点指向上一级节点,上一级节点指向None,重复这个过程,直到递归结束
NewHead = self.reverse_list_recusive(head.next)
head.next.next = head
head.next = None
return NewHead
if __name__ == "__main__":
n1 = ListNode(1)
n2 = ListNode(2)
n3 = ListNode(3)
n4 = ListNode(4)
n1.next = n2
n2.next = n3
n3.next = n4
new_head = Solution().reverse_list(n1)
print(new_head.val)
print(new_head.next.val)
print(new_head.next.next.val)
print(new_head.next.next.next.val)
print()
new_head = Solution().reverse_list2(new_head)
print(new_head.val)
print(new_head.next.val)
print(new_head.next.next.val)
print(new_head.next.next.next.val)
print()
new_head = Solution().reverse_list_recusive(new_head)
print(new_head.val)
print(new_head.next.val)
print(new_head.next.next.val)
print(new_head.next.next.next.val)
THE END
这道题很经常的出现在各个厂的笔试题和面试题中,平时大家估计很少用链表,但是面试经常考,所以也要注意下链表的性质,和怎么反转链表