【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;
}

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