Reorder List -- leetcode

最新推荐文章于 2025-08-08 22:01:10 发布

weixin_30326515

最新推荐文章于 2025-08-08 22:01:10 发布

阅读量52

点赞数

CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/bhlsheji/p/4872909.html

本文介绍了一种链表重组算法，该算法将单链表L从L0→L1→…→Ln-1→Ln重新排列为L0→Ln→L1→Ln-1→L2→Ln-2→…。通过寻找链表中点、反转后半部分链表并最终合并两部分链表实现。在LeetCode上的实际运行时间为72ms。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

基本思路：

先找到中间结点。

然后将后半部分链表进行反转；

再合并这两个链表。

在leetcode上实际运行时间为72ms。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void reorderList(ListNode* head) {
        if (!head || !head->next)
            return;
            
        ListNode *walker = head;
        ListNode *runner = head->next;
        while (runner && runner->next) {
            walker = walker->next;
            runner = runner->next;
            runner = runner->next;
        }
        
        ListNode middle(0);
        runner = walker;
        walker = walker->next;
        runner->next = NULL;
        while (walker) {
            ListNode *bak = walker->next;
            walker->next = middle.next;
            middle.next = walker;
            walker = bak;
        }
        
        while (middle.next) {
            ListNode *bak = middle.next->next;
            middle.next->next = head->next;
            head->next = middle.next;
            head = head->next->next;
            middle.next = bak;
        }
    }
};

转载于:https://www.cnblogs.com/bhlsheji/p/4872909.html