LeetCode 24 Swap Nodes in Pairs (链表递归&&非递归)

最新推荐文章于 2026-01-04 19:40:08 发布

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本文介绍了一种链表操作技巧——成对交换相邻节点的方法。通过两种实现方式：迭代法与递归法，详细解释了如何在常数空间内完成这一任务，而不改变链表中的元素值。

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.

题目链接：https://leetcode.com/problems/swap-nodes-in-pairs/description/

迭代法：四个数字作为一组，在纸上画画即可

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode helper = new ListNode(0);
        helper.next = head.next;
        ListNode pre = head, cur = head, tmp = head;
        while (pre != null && pre.next != null) {
            cur = pre.next;
            tmp = cur;
            cur = cur.next;
            tmp.next = pre;
            if (cur == null) {
                pre.next = null;
                break;
            }
            if (cur.next == null) {
                pre.next = cur;
                cur.next = null;
                break;
            }
            pre.next = cur.next;
            pre = cur;
        }
        return helper.next;
    }
}

递归法：真的好写。。。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode cur = head.next;
        head.next = swapPairs(cur.next);
        cur.next = head;
        return cur;
    }
}