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 oldToNewMap = new HashMap();
        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
}