[Leetcode]#143 Reorder List

最新推荐文章于 2022-10-08 18:49:12 发布

jerseyma

最新推荐文章于 2022-10-08 18:49:12 发布

阅读量366

点赞数

CC 4.0 BY-SA版权

分类专栏： linkedlist 文章标签： leetcode linkedlist

本文链接：https://blog.youkuaiyun.com/jerseyma/article/details/48173461

linkedlist 专栏收录该内容

19 篇文章

订阅专栏

本文介绍了解决LeetCode第143题——重新排列链表的方法，使用双指针技巧实现链表的重新排序。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

https://leetcode.com/problems/reorder-list/

//#143 Reorder List
//64ms 97.92%
/**
 * 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 || !head->next->next) return;
        //size == 0; size == 1; size == 2
        ListNode *p1(head), *p2(head);
        while(p2->next && p2->next->next)
        {
            p1 = p1->next;
            p2 = p2->next->next;    
        }

        if(p1->next && p1->next->next && !p1->next->next->next)
        {
            p1->next->next->next = p1->next;
            p1->next = p1->next->next;
            p1->next->next->next = NULL;
        }
        else if(p1->next && p1->next->next && p1->next->next->next)
        {
            ListNode *previous_p(p1->next), *current_p(p1->next->next), *next_p(p1->next->next->next);
            previous_p->next = NULL;
            while(next_p != NULL)
            {
                current_p->next = previous_p;

                previous_p = current_p;
                current_p = next_p;
                next_p = next_p->next;          
            }
            current_p->next = previous_p;
            p1->next = current_p;
        }

        p2 = head;

        //list_print(head);

        while(p1->next && p1 != p2)
        {
            ListNode *tmp(p2->next);
            p2->next = p1->next;
            p1->next = p1->next->next;
            p2->next->next = tmp;
            //cout << "ok?\n";
            p2 = p2->next->next;
            //cout << "ok?\n";
        }

        return;
    }
};