[leetcode] Swap Nodes in Pairs

最新推荐文章于 2025-09-09 20:07:09 发布

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原文链接：https://my.oschina.net/jdflyfly/blog/284002

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#数据结构与算法 #python

2019独角兽企业重金招聘Python工程师标准>>>

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

For example,
Given1->2->3->4, you should return the list as2->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.

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

思路：模拟题，仔细操作即可。

增加dummyhead方便处理；
操作节点需保留前一个节点的指针。

代码：

public class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null)
            return null;
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        ListNode pre = dummyHead;
        while (pre != null && pre.next != null && pre.next.next != null) {
            ListNode q = pre.next;
            ListNode r = q.next;
            pre.next = r;
            q.next = r.next;
            r.next = q;
            pre = pre.next.next;
        }
        return dummyHead.next;

    }

    public static void main(String[] args) {
        ListNode list = ListUtils.makeList(1, 2, 3, 4, 5);
        ListUtils.printList(list);
        ListUtils.printList(new Solution().swapPairs(list));
    }

}

转载于:https://my.oschina.net/jdflyfly/blog/284002