24 Swap Nodes in Pairs_while(currnext!=null && currnext.next != null){ li-优快云博客

本文链接：https://blog.youkuaiyun.com/dongbeier/article/details/80317549

本文介绍了一种算法，用于交换链表中每两个相邻节点的位置。提供了两种实现方式：递归和迭代。递归方法通过调整指针指向来完成节点交换；迭代方法则使用虚拟头节点简化边界条件处理。

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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.

递归:

/**
 * 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 n = head.next;
        head.next = swapPairs(head.next.next);
        n.next = head;
        return n;
    }
}

迭代:

/**
 * 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 dummy = new ListNode(0);
        dummy.next = head;
        ListNode curr = dummy;
        
        while (curr.next != null && curr.next.next != null) {
            ListNode first = curr.next;
            ListNode second = curr.next.next;
            
            first.next = second.next;
            curr.next = second;
            second.next = first;
            curr = first;
        }
        return dummy.next;
    }
}