Leetcode - CopyWithRandomList

最新推荐文章于 2022-01-23 21:21:02 发布

转载最新推荐文章于 2022-01-23 21:21:02 发布 · 63 阅读

本文介绍了一种复制带有随机指针的链表的方法。通过迭代并复制原始链表，利用哈希映射存储每个节点的旧地址及其对应的新地址。之后，根据映射更新随机指针。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Algorithm: Iterate and copy the original list first. For the random pointer, just copy the value from the original list first. And use a map to store each node's old address and its corresponding new address. After the iteration, we can replace the value of the random pointer based on the map we get.

Mistakes I make:

(1) Beware when head == null.

(2) Forget during the iteration: node = node.next;

public class Solution {
    public RandomListNode copyRandomList(RandomListNode head) {
        if(head == null)
    		return null;
    	RandomListNode newHead = new RandomListNode(head.label);
    	newHead.random = head.random;
    	RandomListNode node = newHead;
    	Map<RandomListNode,RandomListNode> addrRef = new HashMap<RandomListNode,RandomListNode>();
    	
    	addrRef.put(head, node);
    	head = head.next;
    	
    	while(head != null)
    	{
    		RandomListNode tnode = new RandomListNode(head.label);
    		tnode.random = head.random;
    		node.next = tnode;
    		
        	addrRef.put(head, tnode);

    		head = head.next;
    		node = tnode;
    	}
    	
    	node = newHead;
    	while(node != null)
    	{
    		node.random = addrRef.get(node.random);
    		node = node.next;
    	}
    	
    	return newHead;
    	
    }
}