[LeetCode][C++] #138. Copy List with Random Pointer

[Medium][Question]:

A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null.

Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y.

Return the head of the copied linked list.

The linked list is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:

• val: an integer representing Node.val
• random_index: the index of the node (range from 0 to n-1) that the random pointer points to, or null if it does not point to any node.

Your code will only be given the head of the original linked list.

Example 1:

Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]

Example 2:

Input: head = [[1,1],[2,1]]
Output: [[1,1],[2,1]]

Example 3:

Input: head = [[3,null],[3,0],[3,null]]
Output: [[3,null],[3,0],[3,null]]

Example 4:

Input: head = []
Output: []
Explanation: The given linked list is empty (null pointer), so return null.

Constraints:

• 0 <= n <= 1000
• -10000 <= Node.val <= 10000
• Node.random is null or is pointing to some node in the linked list.

My Solution[C++]:

[題目講解]：input一串Linked List，有Next指標和Random指標，而題目要回傳的是一個Deep Copy的Linked List

Deep Copy：完全不同記憶體位置的複製。

[Ideas]: 建立一個Unordered_Map，該Map：

・Key為Node指標，指向舊的Linked List的Node；

・Value為Node指標，指向新的Deep Copy的Node。

分為三個步驟

1. 先跑過一次舊的Linked List，利用Map建立舊Node與新Node的Mapping關係。

2. 跑第二次舊的Linked List，把Next與Random兩個指標，連上新的Node。

3. 最後回傳Map[Head]，代表回傳新的Linked List。