题目描述

题目链接：138. Copy List with Random Pointer

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.

例子

例子 1

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

例子 2

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

Constraints

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

解题思路

这道题目思路比较直观，先通过一次遍历新建出一个单链表（此时先不处理随机节点），并通过哈希表储存节点之间的对应关系，第二次遍历时完善随机节点即可，代码如下：

class Solution {
public:
    Node* copyRandomList(Node* head) {
        Node* curr = head;
        Node* dummy = new Node(0);
        Node* new_curr = dummy;
        while (curr) {
            Node* new_node = new Node(curr->val);
            new_curr->next = new_node;
            hmap[curr] = new_node;
            new_curr = new_curr->next;
            curr = curr->next;
        }

        curr = head;
        while (curr) {
            hmap[curr]->random = hmap[curr->random];
            curr = curr->next;
        }
        return dummy->next;
    }

private:
    std::unordered_map<Node*, Node*> hmap;
};

• 时间复杂度: O(n)
• 空间复杂度: O(n)

GitHub 代码同步地址： 138.CopyListWithRandomPointer.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions