【LeetCode & 剑指offer刷题】链表题8：35 复杂链表的复制（138. Copy List with Random Pointer）

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138. 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.

 

/**

 * Definition for singly-linked list with a random pointer.

 * struct RandomListNode {

 *     int label;

 *     RandomListNode *next, *random;

 *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {}

 * };

 */

/*

问题：复制带随机指针的链表

方法：在原链表内部复制链表结点，由内部关系，复制随机指针连接关系，再进行拆分

*/

class Solution

{

public:

    RandomListNode *copyRandomList(RandomListNode *head)

    {

        //第一步：复制结点与next指针

        RandomListNode* cur = head;

        while(cur != nullptr)

        {

            RandomListNode* node = new RandomListNode(cur->label); //构造函数会将next和random指针赋值为null

            node->next = cur->next;//新结点与旧结点连接 

           

            cur->next = node; //旧结点与新结点连接

            cur = node->next; //下个旧结点

        }

       

        //第二步：复制random指针

        cur = head;

        while(cur != nullptr)

        {

            if(cur->random != nullptr) cur->next->random = cur->random->next; //复制random指针 (cur->next为新结点)

            cur = cur->next->next; //下个旧结点

        }

       

        //第三步：拆分链表

        cur = head;

        RandomListNode prehead(0); //创建整个链表的空头结点,方便处理

        prehead.next = head;

        RandomListNode* newcur = &prehead;

        while(cur != nullptr)

        {

            newcur->next = newcur->next->next;//先连接前面的指针，类似题目 328. Odd Even Linked List

            cur->next = cur->next->next;

            

            newcur = newcur->next; //更新指针

            cur = cur->next; //cur走在newcur前面，方便判断  

        }

        return prehead.next;

    }

};