LeetCode 分类刷题：24. 两两交换链表中的节点

题目

给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。

示例 2：

输入：head = []
输出：[]

示例 3：

输入：head = [1]
输出：[1]

解析

1.参考 LeetCode 分类刷题：25. K 个一组翻转链表

这里相当于k = 2

2.灵神方法

答案

1.

# 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: Optional[ListNode]) -> Optional[ListNode]:
        p0 = dummy = ListNode(next = head)
        cur = head
        n = 0
        while cur:
            n += 1
            cur = cur.next
        
        pre = None
        cur = head
        while n >= 2:
            n -= 2
            for _ in range(2):
                nxt = cur.next
                cur.next = pre
                pre = cur
                cur = nxt
            
            nxt = p0.next
            p0.next.next = cur
            p0.next = pre
            p0 = nxt
        return dummy.next    

2.

# 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: Optional[ListNode]) -> Optional[ListNode]:
        dummy = n0 = ListNode(next = head)
        n1 = head
        while n1 and n1.next:
            n2 = n1.next
            n3 = n2.next

            n0.next = n2
            n2.next = n1
            n1.next = n3

            n0 = n1
            n1 = n3
        return dummy.next

复杂度分析

时间复杂度：O(n)

空间复杂度：O(1)