【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;
}