[数据结构与算法]24. 两两交换链表中的节点（leetcode)

题目分析：
交换相邻节点的顺序，就是改变next指针即可
用个dummy + while来操作一下

# 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:
        dummy = ListNode(0, head)
        cur = dummy
        while cur.next != None and cur.next.next != None:
            p = cur.next
            q = cur.next.next
            p.next = q.next
            q.next = p
            cur.next = q
            cur = cur.next.next
        return dummy.next

题目总结：
用dummy+交换一下next