反转链表(python)
时间: 2023-06-01 07:03:24 浏览: 81
# 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:
prev = None
curr = head
while curr:
temp = curr.next
curr.next = prev
prev = curr
curr = temp
return prev
相关问题
反转链表python
可以使用迭代的方法来反转一个链表。具体的实现代码如下:
```python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def reverseList(head):
prev = None
curr = head
while curr:
next_node = curr.next
curr.next = prev
prev = curr
curr = next_node
return prev
```
这段代码中,我们使用了两个指针 `prev` 和 `curr` 来遍历链表,同时调整指针的指向,实现链表的反转。最后返回 `prev` 指针,它指向了反转后的链表的头节点。
1025 反转链表python
1025 反转链表的Python代码如下所示:
```python
# 开始的地址,链表节点数量,K
start_address, count, k = list(map(int, input().split(' ')))
# 创建一个长度为10万的空列表
single_lists = [None * 100000
# 读取输入,并存放至空列表中
for i in range(count):
single_list = input().split(' ')
single_lists[int(single_list = reverse_result_lists[::-1]
# 按要求输出单链表
for j in range(count - 1):
result_lists[j][2 = result_lists[j + 1][0]
print(' '.join(result_lists[j]))
result_lists[-1][2 = '-1'
print(' '.join(result_lists[-1]))
```
以上代码实现了反转链表的功能,通过读取输入的链表节点,并按照题目要求进行反转和输出。<span class="em">1</span><span class="em">2</span><span class="em">3</span>
#### 引用[.reference_title]
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