Problem: 138. 随机链表的复制

题目描述

思路及解法

1.创建Map集合Map<Node, Node> map;创建指针cur指向head；
2.遍历链表将cur作为键，new Node(cur.val)作为值，存入map集合；
3.再次遍历链表，利用map集合存贮的键，将创建的节点（map集合中的值）连接起来，最后返回新链表的头节点

复杂度

时间复杂度:

$O(n)$；其中$n$为链表的大小

空间复杂度:

$O(n)$

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 {
    /**
     * Copy List with Random Pointer
     *
     * @param head The head of linked list
     * @return Node
     */
    public Node copyRandomList(Node head) {
        if (head == null) {
            return null;
        }
        Node cur = head;
        Map<Node, Node> map = new HashMap<>();
        while (cur != null) {
            map.put(cur, new Node(cur.val));
            cur = cur.next;
        }
        cur = head;
        while (cur != null) {
            map.get(cur).next = map.get(cur.next);
            map.get(cur).random = map.get(cur.random);
            cur = cur.next;
        }
        return map.get(head);
    }
}