链表调试示例 -24.两两交换链表中的节点(中等)

我是胖虎啊

发布于 2022-06-27 18:03:38

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发布于 2022-06-27 18:03:38

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24. 两两交换链表中的节点(中等)

题目链接

https://leetcode-cn.com/problems/swap-nodes-in-pairs/

解题思路

定义一个虚拟的头结点, 辅助遍历
判断当前结点的next结点和next.next结点是否为空, 不为空则互相交换, 为空时, 则输出虚拟结点.next 即为所求!

题解代码

class Solution:
    def swapPairs(self, head: ListNode) -> ListNode:
        dummy = ListNode()
        dummy.next = head
        dummy_tmp = dummy
        while dummy_tmp.next and dummy_tmp.next.next:
            node1 = dummy_tmp.next
            node2 = dummy_tmp.next.next
            dummy_tmp.next = node2
            node1.next = node2.next
            node2.next = node1

            dummy_tmp = node1
        return dummy.next

调试技巧

完整的调试代码

from leetcode.linked_struct import ListNode
from leetcode.linked_struct import SingleLinkList

link_list = SingleLinkList()

node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)
node4 = ListNode(4)

link_list.head = node1
node1.next = node2
node2.next = node3
node3.next = node4

class Solution:
    def swapPairs(self, head: ListNode) -> ListNode:
        dummy = ListNode()
        dummy.next = head
        dummy_tmp = dummy
        while dummy_tmp.next and dummy_tmp.next.next:
            node1 = dummy_tmp.next
            node2 = dummy_tmp.next.next
            dummy_tmp.next = node2
            node1.next = node2.next
            node2.next = node1

            dummy_tmp = node1
        return dummy.next


if __name__ == '__main__':
    link_list.view_linklist()
    a = Solution()
    link_list.head = a.swapPairs(link_list.head)
    link_list.view_linklist()