Leetcode: Copy List with Random Pointer

最新推荐文章于 2022-11-20 16:05:29 发布

原创最新推荐文章于 2022-11-20 16:05:29 发布 · 3k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Leetcode #Copy List with Rando

算法与数据结构同时被 3 个专栏收录

223 篇文章

订阅专栏

程序员笔试面试

223 篇文章

订阅专栏

Leetcode

179 篇文章

订阅专栏

本文介绍了一种复杂链表结构的深拷贝方法，该链表除了常规的next指针外还包含了一个指向链表中任意节点或null的随机指针。通过三个关键步骤：复制链表、设置随机指针以及分离链表，实现了对原链表的完全复制。

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

/**
 * Definition for singly-linked list with a random pointer.
 * struct RandomListNode {
 *     int label;
 *     RandomListNode *next, *random;
 *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
 * };
 */
class Solution {
public:
    void duplicateList(RandomListNode* head)
	{
		if(head==NULL)return;
		RandomListNode* p = head;
		while(p)
		{
			RandomListNode* tmp = new RandomListNode(p->label);
			tmp->next = p->next;
			p->next = tmp;
			p = tmp->next;
		}
	}
	void setRandom(RandomListNode* head)
	{
		if(head==NULL)return;
		RandomListNode* p = head;
		RandomListNode* cpy = p->next;
		while(p)
		{
			if(p->random)
				cpy->random = p->random->next;
			else
				cpy->random = NULL;

			p = cpy->next;
			if(p)
				cpy = p->next;
		}
	}
	RandomListNode* separateList(RandomListNode* head)
	{
		if(head==NULL)return head;
		RandomListNode* phead = head->next;
		RandomListNode* p = head;
		RandomListNode* p1 = phead;
		while(p)
		{
			p->next = p1->next;
			p = p->next;
			if(p){
				p1->next = p->next;
				p1 = p1->next;
			}
		}
		return phead;
	}
	RandomListNode *copyRandomList(RandomListNode *head) {
        // Note: The Solution object is instantiated only once.
        duplicateList(head);
		setRandom(head);
		return separateList(head);
    }
};