leetcode138. Copy List with Random Pointer【剑指offer】

最新推荐文章于 2024-11-27 11:42:05 发布

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最新推荐文章于 2024-11-27 11:42:05 发布

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分类专栏： leetcode 文章标签： leetcode

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本文介绍了一种解决LeetCode上复杂链表深拷贝问题的方法。链表中每个节点除了包含指向下一个节点的指针外，还包含一个随机指针，该指针可以指向链表中的任意节点或空。文章详细阐述了如何在不破坏原链表的情况下，创建一个完全相同的链表副本，包括所有节点及其随机指针指向。

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https://leetcode.com/problems/copy-list-with-random-pointer/

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.

Example 1:

Input:
{"$id":"1","next":{"$id":"2","next":null,"random":{"$ref":"2"},"val":2},"random":{"$ref":"2"},"val":1}

Explanation:
Node 1's value is 1, both of its next and random pointer points to Node 2.
Node 2's value is 2, its next pointer points to null and its random pointer points to itself.

Note:

You must return the copy of the given head as a reference to the cloned list.

leetcode会员刷的第一个

复制复杂链表

吃晚饭的时候想的思路对，之前那遍剑指没白刷

但是忽略一个问题

在拆链表的时候，之前的那个也要复原，不能光拿出来一个新的就完事了

class Solution {
public:
    Node* copyRandomList(Node* head) {
        Node* head0 = head;
        if (head == NULL) {
            return head0;
        }
        while(head0 != NULL) {
            Node *t = new Node();
            t->val = head0->val;
            t->next = head0->next;
            head0->next = t;
            head0 = t->next;
        }
        head0 = head;
        while(head0 != NULL) {
            Node* x = head0 -> random;
            Node* k0 = head0 -> next;
            if (x != NULL)
                k0 -> random = x -> next;
            else
                k0 -> random = NULL;
            head0 = head0 -> next -> next;
        }
        head0 = head -> next;
        Node* head1 = head;
        Node* ans = head -> next;
        while(head != NULL && head0 ->next != NULL) {
            head1 -> next = head1 -> next -> next;
            head0 -> next = head0 -> next -> next;
            head0 = head0 -> next;
        }
        head1->next = NULL;
        return ans;
    }
};