LeetCode：24. Swap Nodes in Pairs（Medium）

1. 原题链接

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

2. 题目要求

给定一个链表，交换相邻的两个结点。已经交换的结点，不再进行交换。

注意：所使用的空间大小固定

例如，1->2->3->4转换后为2->1->4->3

3. 题目思路

使用一个遍历指针current和两个辅助指针first、second，first保存current指针所在结点的后继结点，second保存current指针所在结点的后继的后继结点。

令first的后继指向second的后继，current的后继等于second，current后继的后继等于first，current等于first

4. 代码实现

public class SwapNodeInPairs24 {

    public static void main(String[] args) {

        ListNode l1 = new ListNode(1);

        ListNode l2 = new ListNode(2);

        ListNode l3 = new ListNode(3);

        ListNode l4 = new ListNode(4);

        ListNode l5 = new ListNode(5);

        ListNode l6 = new ListNode(6);

        l1.next = l2;

        l2.next = l3;

        l3.next = l4;

        l4.next = l5;

        l5.next = l6;

        ListNode ls1 = l1;

        while (ls1 != null) {

            System.out.print(ls1.val);

            ls1 = ls1.next;

        }

        System.out.println("");

        ListNode ls2 = swapPairs(l1);

        while (ls2 != null) {

            System.out.print(ls2.val);

            ls2 = ls2.next;

        }

    }

    public static ListNode swapPairs(ListNode head) {

        ListNode headPointer = new ListNode(0);

        headPointer.next = head;

        ListNode current = headPointer;

        while (current.next != null && current.next.next != null) {

            ListNode first = current.next;

            ListNode second = current.next.next;

            first.next = second.next;

            current.next = second;

            current.next.next = first;

            current = first;

        }

        return headPointer.next;

    }

    public static class ListNode {

        int val;

        ListNode next;

        ListNode(int x) {

            val = x;

        }

    }

}