24. Swap Nodes in Pairs Leetcode Python

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本文介绍了一种在常数空间复杂度下交换链表中每两个相邻节点的算法。通过改变节点之间的连接而非节点值来实现节点的交换，并提供了一个包含示例的详细解释。

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

1.先将对应位置的node取出来两者反转

2. 将指针指向p.next.next

1. we need to reverse the two nodes first

2. move pointer to p.next.next

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

class Solution:
    # @param a ListNode
    # @return a ListNode
    def swapPairs(self, head):
        dummy=ListNode(0)
        dummy.next=head
        p=dummy
        while p.next and p.next.next:
            tmp=p.next.next
            p.next.next=tmp.next
            tmp.next=p.next
            p.next=tmp
            p=p.next.next
        return dummy.next