复杂链表的复制

package demo;

/*
 * 题目描述
输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），
返回结果为复制后复杂链表的head。（注意，输出结果中请不要返回参数中的节点引用，否则判题程序会直接返回空）
 */
/*
public class RandomListNode {
    int label;
    RandomListNode next = null;
    RandomListNode random = null;

    RandomListNode(int label) {
        this.label = label;
    }
}
*/
public class Offer_13 {
    public RandomListNode Clone(RandomListNode pHead) {
        RandomListNode p = pHead;
        RandomListNode t = pHead;
        while (p != null) {
            RandomListNode q = new RandomListNode(p.label);
            q.next = p.next;
            p.next = q;
            p = q.next;
        }
        while (t != null) {
            RandomListNode q = t.next;
            if (t.random != null)
                q.random = t.random.next;
            t = q.next;
        }
        RandomListNode s = new RandomListNode(0);
        RandomListNode s1 = s;
        while (pHead != null) {
            RandomListNode r = pHead.next;
            pHead.next = r.next;
            r.next = s.next;
            s.next = r;
            s = s.next;
            pHead = pHead.next;
        }
        return s1.next;
    }
}

class RandomListNode {
    int label;
    RandomListNode next = null;
    RandomListNode random = null;

    RandomListNode(int label) {
        this.label = label;
    }
}