【leetcode】Copy List with Random Pointer (hard)

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.

 

思路：

做过，先复制一遍指针，再复制random位置，再拆分两个链表。

#include <iostream>

#include <vector>

#include <algorithm>

#include <queue>

#include <stack>

using namespace std;





// 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) {}

};

 

class Solution {

public:

    RandomListNode *copyRandomList(RandomListNode *head) {

        if(head == NULL) return NULL;



        RandomListNode * p = head;



        //在每个结点后面复制一个自己 不管random指针

        while(p != NULL)

        {

            RandomListNode * cpy = new RandomListNode(p->label);

            cpy->next = p->next;

            p->next = cpy;



            p = cpy->next;

        }



        //复制random指针

        p = head;

        while(p != NULL)

        {

            RandomListNode * cpy = p->next;

            if(p->random != NULL)

            {

                cpy->random = p->random->next;

            }



            p = cpy->next;

        }



        //把原链表与复制链表拆开

        RandomListNode * cpyhead = head->next;

        p = head;

        RandomListNode * cpy = cpyhead;

        while(p != NULL)

        {

            p->next = cpy->next;

            cpy->next = (cpy->next == NULL) ? cpy->next : cpy->next->next;



            p = p->next;

            cpy = cpy->next;

        }



        return cpyhead;

    }

};



int main()

{

    RandomListNode * r = new RandomListNode(-1);

    Solution s;

    RandomListNode * ans = s.copyRandomList(r);



    return 0;

}