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
}