LeetCode Copy List with Random Pointer
2015-04-20 14:08
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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.思路分析:这题要求拷贝链表,包括内容,next指针和random指针。容易想到的思路是定义一个hashmap保存旧链表节点到新链表节点的映射关系,第一次遍历链表,只拷贝next指针和node,并且init hashmap, 第二遍遍历链表再拷贝所有节点的random指针,借助hashmap可以在O(1)时间内由给定旧链表节点找到新链表对应节点。时间复杂度和空间复杂度都是O(n)。另有一个解法不需要额外空间,需要三遍遍历,要仔细画图分析,详细可见http://codeganker.blogspot.com/2014/03/copy-list-with-random-pointer-leetcode.html http://blog.csdn.net/fightforyourdream/article/details/16879561 AC Code
/** * 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) { //0133 if(head == null) return null; RandomListNode newHead = new RandomListNode(head.label); HashMap<RandomListNode, RandomListNode> oldToNewMap = new HashMap<RandomListNode, RandomListNode>(); oldToNewMap.put(head, newHead); RandomListNode newCur = newHead; // iter for newList RandomListNode cur = head.next; // iter for oriList while(cur != null){ RandomListNode newNode = new RandomListNode(cur.label); newCur.next = newNode; oldToNewMap.put(cur, newNode); newCur = newNode; cur = cur.next; } newCur = newHead; cur = head; while(cur != null){ newCur.random = oldToNewMap.get(cur.random); cur = cur.next; newCur = newCur.next; } return newHead; } //0202 }
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