LeetCode - 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.
如果要copy一个带有random pointer的list,主要的问题就是有可能这个random指向的位置还没有被copy到,所以解决方法都是多次扫描list。
第一种方法,就是使用HashMap来坐,HashMap的key存原始pointer,value存新的pointer。
第一遍,先不copy random的值,只copy数值建立好新的链表。并把新旧pointer存在HashMap中。
第二遍,遍历旧表,复制random的值,因为第一遍已经把链表复制好了并且也存在HashMap里了,所以只需从HashMap中,把当前旧的node.random作为key值,得到新的value的值,并把其赋给新node.random就好。
/** * Definition for singly-linked list with a random pointer. * class RandomListNode { * int label; * RandomListNode next, random; * RandomListNode(int x) { this.label = x; } * }; */ public class Solution { public RandomListNode copyRandomList(RandomListNode head) { if(head == null){ return null; } Map<RandomListNode, RandomListNode> map = new HashMap<>(); RandomListNode oldHead = head; RandomListNode dummy = new RandomListNode(-1); RandomListNode cur = dummy; while(oldHead != null){ RandomListNode newHead = new RandomListNode(oldHead.label); map.put(oldHead, newHead); cur.next = newHead; oldHead = oldHead.next; cur = cur.next; } oldHead = head; cur = dummy.next; while(oldHead != null){ if(map.containsKey(oldHead.random)){ cur.random = map.get(oldHead.random); } else{ cur.random = null; } oldHead = oldHead.next; cur = cur.next; } return dummy.next; } }
posted on 2018-10-02 11:52 IncredibleThings 阅读(77) 评论(0) 编辑 收藏 举报