138. Copy List with Random Pointer**

138. Copy List with Random Pointer**

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

题目描述

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.

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​​​) where random pointer points to, or​​null​​ if it does not point to any node.

Example 1:

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

Example 2:

Input: head = [[1,1],[2,1]]
Output: [[1,1],[2,1]]

Example 3:

Input: head = [[3,null],[3,0],[3,null]]
Output: [[3,null],[3,0],[3,null]]

Example 4:

Input: head = []
Output: []
Explanation: Given linked list is empty (null pointer), so return null.

Constraints:

• ​​-10000 <= Node.val <= 10000​​
• ​​Node.random​​​ is​​null​​ or pointing to a node in the linked list.
• Number of Nodes will not exceed 1000.

C++ 实现 1

这道题感觉做了好多遍, 好像 “剑指Offer” 上也有这道题. 思路是: 访问链表中的每个节点 ​​root​​​, 产生一个值相同的新节点 ​​p​​​, 同时保存 ​​root -> p​​​ 的映射关系, 在这一步不考虑 ​​random​​​, 只考虑 ​​next​​​. 这样就可以重建一个新的链表, 但 ​​random​​ 还没有指向正确的节点.

下一步遍历新链表, 通过映射关系更新新链表中的 ​​random​​ 的最终指向.

/*
// 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) return nullptr;
        unordered_map<Node*, Node*> record;
        auto root = head;
        auto dummy = new Node(0);
        auto p = dummy;
        while (root) {
            p->next = new Node(root->val);
            p = p->next;
            record[root] = p;
            root = root->next;
        }
        p = dummy->next;
        root = head;
        while (p) {
            p->random = record[root->random];
            root = root->next;
            p = p->next;
        }
        return dummy->next;
    }
};