python leetcode编程作业3

最新推荐文章于 2025-07-12 17:40:17 发布

原创最新推荐文章于 2025-07-12 17:40:17 发布 · 131 阅读

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本文介绍了一种链表操作算法——成对交换相邻节点。通过仅使用常数额外空间，实现了链表中节点值的有效交换，而无需改变节点本身。此算法适用于需要对链表进行快速调整的应用场景。

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题目：Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

Note:

Your algorithm should use only constant extra space.
You may not modify the values in the list's nodes, only nodes itself may be changed.

解题思路：

从链表头开始，每两个之间进行值的交换，然后切换到后两个继续交换直到为空

代码：

class ListNode:
     def __init__(self, x):
         self.val = x
         self.next = None

class Solution:
    def swapPairs(self, head):
        if not head:
            return head
        first = head
        second = first.next
        while first and second:
            change = first.val
            first.val = second.val
            second.val = change
            if second.next:
                first = second.next
                second = first.next
            else:
                break
        return head