[LeetCode]24. Swap Nodes in Pairs

最新推荐文章于 2025-08-20 09:37:41 发布

转载最新推荐文章于 2025-08-20 09:37:41 发布 · 191 阅读

文章标签：

#leetcode

leetcode 专栏收录该内容

64 篇文章

订阅专栏

本文介绍了一种在常数空间复杂度下交换链表中每两个相邻节点的算法。通过迭代方式实现，不修改节点值仅调整节点指向，最终返回交换后的链表头节点。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Easy

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

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

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

0ms:

  public ListNode swapPairs(ListNode head) {
       if(head==null||head.next==null)  return head;
         ListNode root = new ListNode(0);
         root.next = head;
         ListNode node = root;
         ListNode tmp = root.next;

         while(tmp!=null&&tmp.next!=null){
             node.next = tmp.next;
             tmp.next = tmp.next.next;
             node.next.next = tmp;
             node = tmp;
             tmp = node.next;
         }
         return root.next;

    }

0ms:

 public ListNode swapPairs(ListNode head) {
        if(head==null||head.next==null)  return head;

       ListNode root = new ListNode(0);
         root.next = head;
         ListNode node = root;
         ListNode tmp = root.next;
         ListNode root2 = tmp.next;
         while(tmp!=null&&tmp.next!=null){
             node.next = tmp.next;
             tmp.next = tmp.next.next;
             node.next.next = tmp;
             node = tmp;
             tmp = node.next;
         }
         return root2;

    }