36：Copy List with Random Pointer

最新推荐文章于 2025-05-21 11:23:47 发布

weishenmetlc

最新推荐文章于 2025-05-21 11:23:47 发布

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分类专栏： leetcode

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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。该链表每个节点除了包含指向下一个节点的指针外，还包含一个指向链表中任意节点的随机指针。通过巧妙地插入和拆分节点的方式，实现了一个高效的解决方案。

本题解法代码的思想及编写参考了网址https://github.com/soulmachine/leetcode#leetcode题解

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.

解题代码如下：

class Solution {
public:
        RandomListNode* copyRandomList(RandomListNode* head) {
                for (RandomListNode* cur = head; cur != nullptr;) {
                        RandomListNode* node = new RandomListNode(cur -> label);                        
                        node -> next = cur -> next;
                        cur -> next = node;
                        cur = node -> next;
                }

                for (RandomListNode* cur = head; cur != nullptr; ) {
                        if (cur -> random != nullptr)
                                cur -> next -> random = cur -> random -> next;
                        cur = cur -> next -> next;
                }

                // 分拆成两个单链表
                RandomListNode dummy{-1};
                for (RandomListNode *cur = head, *new_cur = &dummy; cur != nullptr; ) {
                        new_cur -> next = cur -> next;
                        new_cur = new_cur -> next;
                        cur -> next = cur -> next -> next;
                        cur = cur -> next;
                }
                return dummy.next;
        }
};