# 面试题35

请实现 copyRandomList 函数，复制一个复杂链表。在复杂链表中，每个节点除了有一个 next 指针指向下一个节点，还有一个 random 指针指向链表中的任意节点或者 null。

```
示例 1：

输入：head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
输出：[[7,null],[13,0],[11,4],[10,2],[1,0]]

示例 2：

输入：head = [[1,1],[2,1]]
输出：[[1,1],[2,1]]

示例 3：

输入：head = [[3,null],[3,0],[3,null]]
输出：[[3,null],[3,0],[3,null]]

示例 4：

输入：head = []
输出：[]
解释：给定的链表为空（空指针），因此返回 null。
```

## 提示：

* -10000 <= Node.val <= 10000
* Node.random 为空（null）或指向链表中的节点。
* 节点数目不超过 1000 。

## 注意：本题与主站 138 题相同：[https://leetcode-cn.com/problems/copy-list-with-random-pointer/](https://leetcode-cn.com/problems/copy-list-with-random-pointer/)

## Solutions

1. **hashtable**
2. Create a unique map of nodes in the original list and nodes in new list.

```cpp
/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* next;
    Node* random;

    Node(int _val) {
        val = _val;
        next = NULL;
        random = NULL;
    }
};
*/
class Solution {
public:
    Node * node(unordered_map<Node *, Node *> & m, Node * src) {
        if (!src) return nullptr;
        if (!m.count(src))
            m[src] = new Node(src->val);
        return m[src];
    }

    Node* copyRandomList(Node* head) {
        unordered_map<Node *, Node *> m;
        Node * cur = head;

        while (cur) {
            Node * newhead = node(m, cur);
            newhead->next = node(m, cur->next);
            newhead->random = node(m, cur->random);
            cur = cur->next;
        }

        return m[head];
    }
};
```

1.  **inplace hashtable**

    ```cpp
    class Solution {
    public:
     Node* copyRandomList(Node* head) {
         if (!head) return nullptr;
         Node * cur = head;
         while (cur) {
             Node * nnext = cur->next;
             cur->next = new Node(cur->val);
             cur = cur->next->next = nnext;
         }

         cur = head;
         while (cur) {
             if (cur->random)
                 cur->next->random = cur->random->next;
             cur = cur->next->next;
         }

         cur = head;
         Node * newhead = head->next;
         while (cur) {
             Node * nnext = cur->next->next;
             if (nnext)
                 cur->next->next = nnext->next;
             cur = cur->next = nnext;
         }

         return newhead;
     }
    };
    ```
