leetcode 143 —— Reorder List

最新推荐文章于 2022-05-09 11:45:07 发布

原创最新推荐文章于 2022-05-09 11:45:07 发布 · 491 阅读

0 ·

CC 4.0 BY-SA版权

算法学习同时被 2 个专栏收录

166 篇文章

订阅专栏

数据结构

163 篇文章

订阅专栏

本文介绍了一种链表重组算法，该算法将一个单链表重新排序为特定模式，即首尾相接交替的方式。通过计算链表长度并将其分为两部分，接着反转后半部分链表，最后交错合并两部分链表实现目标。整个过程不改变节点值且在原地完成。

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}.

思路：

class Solution {
public:
	void reorderList(ListNode* head) {
		if (!head || !head->next) return;
		int len = 0;
		ListNode* end2 = head;
		while (end2->next){                 //计算长度
			len++;
			end2 = end2->next;
		}
																	
		len = len/ 2;				//分离	
		ListNode *head2 = head;
		while (len){
			head2 = head2->next;
			len--;
		}
		ListNode* end1 = head2;
		head2 = head2->next;
		end1->next = nullptr;

		ListNode* cur = head2;              //后半段反序
		ListNode* nex = cur->next;
		head2->next = nullptr;
		while (nex){
			cur = nex;
			nex = nex->next;
			cur->next = head2;
			head2 = cur;
		}

		ListNode* list1 = head;              //排列
		ListNode* next1 = list1->next;
		ListNode* list2 = head2;
		ListNode* next2 = list2->next;
		while (list1&&list2){
			list1->next = list2;
			list2->next = next1;
			list1 = next1;
			list2 = next2;
			if (next1)
				next1 = next1->next;
			if (next2)
				next2 = next2->next;
		}
		return;
	}
};