138. Copy List with Random Pointer

Problem:

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.

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) where random pointer points to, or null if it does not point to any node.

 

Example 1:

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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: Given linked list is empty (null pointer), so return null.

 

Constraints:

• -10000 <= Node.val <= 10000
• Node.random is null or pointing to a node in the linked list.
• Number of Nodes will not exceed 1000.

 

Analysis:

本题的思路使用Map来映射旧节点和新节点，然后再遍历一遍链表组装新的链表结构。代码如下：

Code:

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

    public Node(int val) {
        this.val = val;
        this.next = null;
        this.random = null;
    }
}
*/
class Solution {
    public Node copyRandomList(Node head) {
        Map<Node, Node> nodes = new HashMap<Node, Node>();
        Node p = head;
        while(p != null) {
            nodes.put(p, new Node(p.val));
            p = p.next;
        }
        
        Node newNode = new Node(-1);
        Node curr = newNode;
        p = head;
        while(p != null) {
            Node temp = nodes.get(p);
            Node newRandom = nodes.get(p.random);
            temp.random = newRandom;
            curr.next = temp;
            curr = curr.next;
            p = p.next;
        }
        return newNode.next;
    }
}