Leetcode 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.

随机链表的节点数据结构

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

注意本题是对带有随机链表指针的深度拷贝，

如果数据结构中那个没有随机指针，只需要将链表遍历拷贝一遍即可

但题目给的数据结构含有指针，拷贝后就必须保有随机指针的关系

本题通过在每个节点之后插入一个当前节点的拷贝节点，然后让插入的节点保有当前节点随机指针的关系，最后将插入的节点从链表中剥离出来即可

主要分为3步

（1）插入拷贝的节点

（2）复制random指针

（3）将插入的节点分离出来

/**
 * 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) {
        //copy list
        RandomListNode *p = head;
        while(p){
            RandomListNode *tmp = new RandomListNode(p->label);
            tmp->next = p->next;
            p->next = tmp;
            p = tmp->next;
        }
        //copy random pointer
        RandomListNode *q = head;
        while(q){
            RandomListNode *tmp = q->next;
            if(q->random){
                tmp->random = q->random->next;
            }
            q=tmp->next;
        }
        //seperate list
        RandomListNode *dupHead = head == NULL ? NULL : head->next, *cur = head;
        while(cur){
            RandomListNode *tmp = cur->next;
            cur->next = tmp->next;
            if(tmp->next){
                tmp->next = tmp->next->next;
            }
            cur = cur->next;
        }
        return dupHead;
    }
};