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 1:
输入:head = [1,2,3,4]
输出:[2,1,4,3]
示例 2:
输入:head = []
输出:[]
示例 3:
输入:head = [1]
输出:[1]
自己写的不对,还是没有理解链表
class Solution:
def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
n = 0
while head:
head = head.next
n += 1
if n < 2:
return head
dummy = ListNode(0)
dummy.next = head
cur = dummy
for _ in range(n//2 + 1):
cur.next = cur.next.next.next
cur.next.next = cur.next.next.next.next
return dummy
官方解
# 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 = ListNode(0)
dummy.next = head
cur = dummy
while cur.next and cur.next.next:
a, b = cur.next, cur.next.next
cur.next, a.next = b, b.next
b.next = a
cur = cur.next.next
return dummy.next