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# Leetcode 138 Copy List with Random Pointer

## Problem

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

Success Details >
Runtime: 28 ms, faster than 96.65% of Python3 online submissions for Copy List with Random Pointer.
Memory Usage: 14.9 MB, less than 69.79% of Python3 online submissions for Copy List with Random Pointer.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28  """ # Definition for a Node. class Node: def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None): self.val = int(x) self.next = next self.random = random """ class Solution: def copyRandomList(self, head: 'Node') -> 'Node': v, r = [], [] d = {} i = 0 node = head while node: v += [node.val] r += [node.random] d[node] = i i, node = i+1, node.next r = [d[node] if node else -1 for node in r] re = [Node(_) for _ in v] + [None] for i in range(len(re)-1): re[i].next = re[i+1] re[i].random = re[r[i]] return re 

## dict.get() 解法

• 使用 get() 很好的避免沒有 key 的狀況，get() 會自動 return None
• 明暸的 two pass
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22  """ # Definition for a Node. class Node: def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None): self.val = int(x) self.next = next self.random = random """ class Solution: def copyRandomList(self, head: 'Node') -> 'Node': d = {} node = head while node: d[node] = Node(node.val) node = node.next node = head while node: d[node].next = d.get(node.next) d[node].random = d.get(node.random) node = node.next return d.get(head) 

## 延伸主題

• defaultdict
• __getitem__
• __missing__