LeetCode Copy List with Random Pointer

最新推荐文章于 2019-10-21 22:50:54 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。这种链表的节点包含一个额外的随机指针，指向链表中的任意节点或空。通过巧妙地在原链表中插入新节点并调整指针，最终实现新旧链表的分离。

题目：

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:
	RandomListNode *copyRandomList(RandomListNode *head) {
		if (head == NULL)
			return head;
		RandomListNode *p = head;
		//对每个节点进行复制，把新节点接在旧节点的下一个
		while (p != NULL) {
			RandomListNode *newNode = new RandomListNode(p->label);
			newNode->next = p->next;
			p->next = newNode;
			p = newNode->next;
		}
		p = head;
		//新节点的random指针为旧节点random指针的下一个或空
		while (p != NULL) {
			RandomListNode *newNode = p->next;
			if(p->random != NULL) 
			    newNode->random = p->random->next;
			p = newNode->next;
		}
		p = head;
		RandomListNode *newHead = head->next;
		//将新旧链表拆分
		while (p != NULL) {
			RandomListNode *newNode = p->next;
			p->next = newNode->next;
			if (newNode->next != NULL) 
				newNode->next = newNode->next->next;
			p = p->next;
		}
		return newHead;
	}
};