Copy List with Random Pointer

最新推荐文章于 2025-05-21 11:23:47 发布

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本文介绍了一种解决复杂链表深拷贝问题的方法，该链表中的节点包含一个额外的随机指针。文章详细阐述了通过构建原始节点与新节点之间的映射来处理随机指针的方法。

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.

解题思路:简单的深拷贝思想，重点在于如何处理random结点,我用的是建立一个原本结点和新建结点的映射.

#include<iostream>
#include<vector>
#include<map>
using namespace std;

//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) {}
};

RandomListNode *copyRandomList(RandomListNode *head) {

	RandomListNode* TmpHead = head;
	RandomListNode* ResultList = new RandomListNode(0);
	RandomListNode* TmpResultNode = ResultList;

	map<RandomListNode*, RandomListNode*>NodeMap;
	while (TmpHead!=NULL)
	{
		RandomListNode* p = new RandomListNode(TmpHead->label);
		NodeMap.insert(make_pair(TmpHead, p));
		TmpResultNode->next = p;
		TmpResultNode = TmpResultNode->next;
		TmpHead = TmpHead->next;
	}
	TmpHead = head;
	TmpResultNode = ResultList->next;
	while (TmpHead!=NULL)
	{
		if (TmpHead->random != NULL)
			TmpResultNode->random = NodeMap[TmpHead->random];
		TmpHead = TmpHead->next;
		TmpResultNode = TmpResultNode->next;
	}
	return ResultList->next;
}