leetcode——138. Copy List with Random Pointer

最新推荐文章于 2025-12-03 23:35:37 发布

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这篇博客讨论了一种复杂的数据结构问题——如何创建一个深度克隆的链表，其中每个节点包含一个额外的随机指针，可能指向链表中的任何节点或空。解决方案涉及两次遍历链表，首先复制节点并调整连接，然后分离原始链表和复制链表。这种方法确保了新链表的随机指针正确地指向新节点，而不是原始链表中的节点。

描述

A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null.

Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y.

Return the head of the copied linked list.

The linked list is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:

val: an integer representing Node.val
random_index: the index of the node (range from 0 to n-1) that the random pointer points to, or null if it does not point to any node.

Your code will only be given the head of the original linked list.

Example 1:

在这里插入图片描述

Input head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* next;
    Node* random;
    
    Node(int _val) {
        val = _val;
        next = NULL;
        random = NULL;
    }
};
*/

class Solution {
public:
    Node* copyRandomList(Node* head) {
        if(head == nullptr)
            return head;
        Node ans(0);
        for(Node *p = head; p!=nullptr;){
            Node *node = new Node(p->val);
            node->next = p->next;
            p->next = node;
            p = node->next;
        }
        for(Node *p = head; p!=nullptr;){
            if(p->random != nullptr)
                p->next->random = p->random->next;
            p = p->next->next;
        }
        ans.next = head->next;
        for(Node *p = head, *cur = head->next; p!=nullptr;){
            p->next = cur->next;
            if(p->next != nullptr)
                cur->next = p->next->next;
            p = p->next;
            cur = cur->next;
        }
        return ans.next;
    }
};