一.题目

 二.思路分析

1.将拷贝节点插入到原节点后面

拷贝节点和原节点建立了一个关联关系

2.深拷贝

3.将新节点拿下来尾插形成新链表，恢复原链表

三.参考代码

/**
 * Definition for a Node.
 * struct Node {
 *     int val;
 *     struct Node *next;
 *     struct Node *random;
 * };
 */

struct Node* copyRandomList(struct Node* head) {
    //指针cur遍历原链表
    struct Node* cur = head;
    while(cur)
    {
        //创建新节点(拷贝节点)，插入到原链表
        struct Node* newnode = (struct Node*)malloc(sizeof(struct Node));
        newnode->val = cur->val;
        newnode->next = cur->next;
        cur->next = newnode;
        cur = newnode->next;
    }
    //重置cur到head，用来重新遍历链表
    cur = head;
    //进行深拷贝random
    while(cur)
    {
        struct Node* copy = cur->next;
        if(cur->random == NULL)
        {
            copy->random = cur->random;//NULL
        }
        else
        {
            copy->random = cur->random->next;//关键句          
        }
        cur = copy->next;
    }
    //重置cur到head
    cur = head;
    //把新节点拿下来尾插成为新链表，并修复原链表
    struct Node* copyhead = NULL,*copytail =NULL;
    while(cur)
    {
        struct Node* copy = cur->next;//每次都是新建立的节点
        struct Node* next = copy->next;//创建指针指向copy的next指针（位于原链表），便于修复原链表
        if(copytail != NULL)
        {
            copytail->next = copy;
            copytail = copytail->next;
        }
        else
        {
            copyhead = copytail = copy;
        }
        cur->next = next;//恢复原链表
        cur = next;//cur移到原链表的下一位    
    }
    return copyhead;
}