Daily question python87: exchange nodes in the linked list in pairs

subject ： I'll give you a list , Two or two exchange the adjacent nodes , And return the head node of the linked list after exchange . You must complete this problem without modifying the value inside the node （ namely , Only node switching can be performed ）.

Example 1：

Input ：head = [1,2,3,4] Output ：[2,1,4,3]
Example 2：

Input ：head = [] Output ：[]
Example 3：

Input ：head = [1] Output ：[1]

Tips ：

The number of nodes in the linked list is in the range [0, 100] Inside 0 <= Node.val <= 100

Program description ： If dummy nodes are not used , The code is like the following error code , If the linked list is disconnected （ According to the code drawing ）

All the code ：
Correct code ：

# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
p = ListNode(0)
p.next = head
cur = p
while cur.next and cur.next.next:
node1 = cur.next
node2 = cur.next.next
cur.next = node2
node1.next = node2.next
node2.next = node1
cur = node1
return p.next

Error code ：

# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
if not head or not head.next:
return head
p = head
cur = head.next
q = cur.next
while cur.next:
p.next = q
cur.next = p
p = q
cur = p.next
q = cur.next
cur.next = p
return head

Title source ： Power button （leetcode）