138. 复制带随机指针的链表

目录

• 题目描述
• 代码实现

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 nullptr.
Return a deep copy of the list.

/**
 * Definition for singly-linked list with a random pointer.
 * struct Node {
 *     int val;
 *     Node *next, *random;
 *     Node(int x) : val(x), next(NULL), random(NULL) {}
 * };
 */

代码实现

class Solution {
public:
    Node *copyRandomList(Node *head) {

        if(head == nullptr)
            return nullptr;
        Node *p = head;
        //第一步
        while(p!=nullptr)
        {
            Node *next = p->next;
            Node *temp = new Node(p->val);
            p->next = temp;
            temp->next = next;
            p = next;
        }
        //第二步
        p = head;
        while(p!=nullptr)
        {   
            //根据程序的逻辑，P不为空，它后面紧跟复制节点p->next
            //一定也不为空
            Node *next = p->next->next;
            if(p->random!=nullptr)
                p->next->random = p->random->next;
            p = next;
        }
        //第三步
        Node *newHead = head->next;
        p = head->next->next;
        //尾指针的运用，由于我们这里没有借用头节点的技巧，所以这里，在进入循环前，先
	//需要把第一个节点处理好，以做为尾节点使用
        Node *last1 = head;
        Node *last2 = head->next;
        last1->next = nullptr;
        last2->next = nullptr;
        while(p!=nullptr)
        {
            Node *next = p->next->next;

            last2->next = p->next;
            last2 = last2->next;
            last2->next = nullptr;

            last1->next = p;
            last1 = last1->next;
            last1->next = nullptr;

            p = next;
        }

        return newHead;       
    }
};