[LeetCode] 重排链表 reorder linked list

最新推荐文章于 2024-04-30 01:45:26 发布

原创最新推荐文章于 2024-04-30 01:45:26 发布 · 816 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

链表专栏收录该内容

22 篇文章

订阅专栏

本文介绍了一种将链表L0→L1→…→Ln-1→Ln重排序为L0→Ln→L1→Ln-1→L2→Ln-2→…的算法实现。通过找到链表中点、反转后半部分链表并插入到前半部分链表的相应位置来完成重排序。提供了完整的C++代码实现。

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}.

Analysis:

Let's see some examples:

{1,2,3,4,5,6} ---> {1,6,2,5,3,4}
{1,2,3,4,5,6,7} ---> {1,7,2,6,3,5,4}

One straightforward middle step of such reordering is:
{1,2,3,4,5,6} --> {1,2,3,6,5,4} --> {1,6,2,5,3,4}
{1,2,3,4,5,6,7}---> {1,2,3,4,7,6,5} ---> {1,7,2,6,3,5,4}

By reversing the last part of the linked list, we do not need to worried about the "parent" pointer anymore. The final step is just insert the each element in the last part into the first part (every two element).

So the algorithm implemented below can be summarized as:
Step 1 Find the middle pointer of the linked list (you can use the slow/fast pointers)
Step 2 Reverse the second part of the linked list (from middle->next to the end)
Step 3 Do the reordering. (inset every element in the second part in between the elements in the first part)

下面是代码。要注意的是：1）第24和25行，fast pointer被初始化成slow pointer的下一个。（自己在纸上画一下，这样初始化的结果是什么）。不能把fast pointer 和slow pointer都初始化成head节点。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
#include <string.h>

#include <stack>
#include <stdio.h>
#include <stdlib.h>

struct ListNode {
     int data;
     ListNode *next;
     ListNode(int x, ListNode* nex) : data(x), next(nex) {}
};

ListNode* reverseList(ListNode* head);
void printLL(ListNode* head);

void reorderList(ListNode *head) {
	if(head==NULL || head->next == NULL || head->next->next==NULL)
		return;

	ListNode* slow = head;
	ListNode* fast = head->next; // fast should be the next of slow

	while(fast)
	{
		if(fast->next)	fast = fast->next;
		else break;

		if(fast->next) fast = fast->next;
		else break;		

		if(slow==NULL) slow = head;
		else slow = slow->next;
	}

	reverseList( slow->next );
	slow->next = NULL;

	ListNode* cur1 = head;
	ListNode* cur2 = fast;
	while(cur1)
	{
		ListNode* nex1 = cur1->next;
		ListNode* nex2 = cur2->next;

		cur1->next = cur2;
		if(nex1 == NULL)
		{
			cur2->next = nex2;
			break;
		}
		else
			cur2->next = nex1;

		cur1 = nex1;
		cur2 = nex2;	
	}	
}

ListNode* reverseList(ListNode* head)
// reverse a linked list
{
	if(head==NULL)
		return head;
 	if(head->next == NULL)
		return head;

	ListNode* tmp = reverseList(head->next);
	tmp->next = head;
	head->next = NULL;
	return head;
}

void printLL(ListNode* head)
{
	while(head)
	{
		cout<<head->data<<" ";
		head = head->next;
	}
}

int main()
{
	ListNode* n0 =new ListNode(0, NULL);
	ListNode* n1 =new ListNode(1, n0);
	ListNode* n2 =new ListNode(2, n1);
	ListNode* n3 =new ListNode(3, n2);
	ListNode* n4 =new ListNode(4, n3);
	ListNode* n5 =new ListNode(5, n4);
	ListNode* n6 =new ListNode(6, n5);
	ListNode* n7 =new ListNode(7, n6);

	reorderList(n7);
	printLL(n7);
}