成对的交换链表的节点 Swap Nodes in Pairs

最新推荐文章于 2025-09-12 17:31:31 发布

转载最新推荐文章于 2025-09-12 17:31:31 发布 · 77 阅读

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原文链接：https://my.oschina.net/liyurong/blog/1527880

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#数据结构与算法

本文提供了一种在常量空间复杂度下交换链表中每两个相邻节点的两种方法。第一种方法通过直接交换节点值来实现，而第二种方法则是通过调整节点链接的方式完成交换，同时确保了不修改节点值。

为什么80%的码农都做不了架构师？>>>

问题：

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.

解决：

① 直接交换即可。

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution { //4ms
public ListNode swapPairs(ListNode head) {
if(head == null || head.next == null) return head;
ListNode slow = head;
ListNode fast = head.next;
while(fast != null && fast.next != null && fast.next.next != null){//避免空指针异常
int tmp = slow.val;
slow.val = fast.val;
fast.val = tmp;
slow = fast.next;
fast = slow.next;
}
int tmp = slow.val;//处理最后两个数
slow.val = fast.val;
fast.val = tmp;
return head;
}
}

②

class Solution {//5ms
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) return head;
ListNode header = new ListNode(-1);
header.next = head;
ListNode cur = header;
ListNode slow = head;
ListNode fast = head.next;
while(slow != null && fast != null){
slow.next = fast.next;
fast.next = slow;
cur.next = fast;
cur = slow;
slow = slow.next;
if(slow != null)fast = slow.next;
}
return header.next;
}
}