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:這題要求拷貝鏈表,包括內容,next指針和random指針。容易想到的思路是定義一個hashmap保存舊鏈表節點到新鏈表節點的映射關係,第一次遍歷鏈表,只拷貝next指針和node,並且init hashmap, 第二遍遍歷鏈表再拷貝所有節點的random指針,藉助hashmap可以在O(1)時間內由給定舊鏈表節點找到新鏈表對應節點。時間複雜度和空間複雜度都是O(n)。
另有一個解法不需要額外空間,需要三遍遍歷,要仔細畫圖分析,詳細可見
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
- }