138. 复制带随机指针的链表

题目出处： https://leetcode-cn.com/problems/copy-list-with-random-pointer/

思路：两次遍历链表。

1. 第一次遍历源链表时创建每个节点的副本，同时将各个副本节点按照next指针相连，并用哈希表记录源节点到对应副本节点的映射。

2. 第二次同时遍历源链表和副本链表，并将源节点随机指针指向的节点的哈希映射节点找出，将该副本节点随机指针指向找出的映射节点，直至遍历完所有节点。

3. 返回副本节点。

/*
// 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) {
        Node* newp = new Node(-1);
        Node* res = newp; 

        Node* p = head; 
        unordered_map<Node*, Node*> mp;
        while(p != nullptr){
            Node* newNode = new Node(p->val);
            newp->next = newNode;
            mp[p] = newNode;
            newp = newp->next; 
            p = p->next;   
        }

        p = head;
        newp = res->next;  
        while(p != nullptr){
            newp->random = mp[p->random];
            p = p->next;
            newp = newp->next; 
        }  

        return res->next; 
    }
};

标签：Node,副本,next,链表,newp,138,节点,指针
来源： https://blog.csdn.net/zaa_12346/article/details/118787976