Copy List with Random Pointer

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题：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.

题意：copy一个链表（深拷贝），链表包含一个next指针和一个random指针，其中random指针指向字符串中的某个节点或者为空。

分析：

第一步: 在原链表的每个节点后面，复制一个新的节点（就是把每个节点复制两份），原链表长度变为 2 倍，random 指针指向的是原链表节点 random 指针指向的节点的后面的那个节点

第二步: 将链表拆成两个 lists

实现：

节点结构：

class RandomNode {
int sign;
RandomNode next, random;
RandomNode(int x) { this.sign= x; }

}

算法代码：

public class leetcode {class RandomNode {    int sign;    RandomNode next, random;    RandomNode(int x) { this.sign = x; }};    public RandomNode copyList(RandomNode head) {        if (head == null) return null;                RandomNode oldnode = head;        while (oldnode != null) {        RandomNode newnode = new RandomNode(oldnode.sign);            newnode.next = oldnode.next;            oldnode.next = newnode;            oldnode = newnode.next;        }                oldnode = head;        while (oldnode != null) {            if (oldnode.random != null) oldnode.next.random = oldnode.random.next;            oldnode = oldnode.next.next;        }                RandomNode newhead = head.next;        oldnode = head;        while (oldnode != null) {        RandomNode newnode = oldnode.next;        oldnode.next = newnode.next;            if (newnode.next != null) newnode.next = newnode.next.next;            oldnode = oldnode.next;        }                return newhead;    }}

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