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);}
}