两两交换链表中的节点 两种解法（Python）

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 = nextclass 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 = nextclass 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

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