Copy List with Random Pointer

CSLDBLYDX

于 2014-12-22 00:17:26 发布

阅读量295

点赞数

CC 4.0 BY-SA版权

分类专栏： leetCode

本文链接：https://blog.youkuaiyun.com/CSLDBLYDX/article/details/42067681

leetCode 专栏收录该内容

39 篇文章

订阅专栏

本文介绍了一种解决复杂链表深拷贝问题的方法。这种链表的节点包含常规的下一个节点指针以及额外的随机指针。文章详细展示了如何通过巧妙地插入新节点并调整指针来完成这一任务。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

class Solution {
public:


    RandomListNode *copyRandomList(RandomListNode *head) {
        
		if(head == NULL)return NULL;
		RandomListNode *p,*newp;
		p = head;
		while(p != NULL)
		{
			newp = new RandomListNode(p->label);
			newp->next = p->next;
			p->next = newp;
			p = p->next->next;
		}

	    RandomListNode *h = head->next;
		newp = h;
		p = head;
		while(1)
		{
			if(p->random)newp->random = p->random->next;
			p = p->next->next;
			if(!p)break;
			newp = newp->next->next;
		}

		newp = h;
		p = head;
		while(1)
		{
			p->next = p->next->next;
			if(newp->next)newp->next = newp->next->next;
			
			p = p->next;
			newp = newp->next;
			if(!newp)break;
			
		}

		return h;
    }
};