[LeetCode 解題報告]138. Copy List with Random Pointer

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:

 

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.

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

    Node() {}

    Node(int _val, Node* _next, Node* _random) {
        val = _val;
        next = _next;
        random = _random;
    }
};
*/
class Solution {
public:
    Node* copyRandomList(Node* head) {
        if (head == NULL)
            return NULL;
        
        // 7->7'->13->13'->11->11'->10->10'->1->1'->null
        Node *cur = head;
        while (cur) {
            Node *node = new Node(cur->val);
            node->next = cur->next;
            cur->next = node;
            cur = node->next;
        }
        
        cur = head;
        while(cur) {
            if (cur->random)
                cur->next->random = cur->random->next;
            cur = cur->next->next;
        } 
        
        cur = head;
        Node *res = head->next;
        while (cur) {
            Node *tmp = cur->next;
            cur->next = tmp->next;
            if(tmp->next)
                tmp->next = tmp->next->next;
            cur = cur->next;
        }
        return res;
    }
};