两两交换链表中的节点 两种解法（Python）

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迭代 时间复杂度 O(N)

# 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_head = ListNode()
        dummy_head.next = head
        tmp = dummy_head
        while tmp.next and tmp.next.next:
            node1 = tmp.next
            node2 = tmp.next.next
            tmp.next = node2
            node1.next = node2.next
            node2.next = node1
            tmp = node1
        return dummy_head.next

递归 时间复杂度O(N)

# 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
        one = head
        two = head.next
        three = head.next.next
        two.next = one
        one.next = self.swapPairs(three)
        return two