[LeetCode] 138. Copy List with Random Pointer java__java

    /**138. Copy List with Random Pointer     * @param head     * @return 深度複製一個鏈表，使用重新分配記憶體儲存新的     */    public RandomListNode copyRandomList(RandomListNode head) {        if (head == null || head.next == null) {            return head;        }        Map<RandomListNode, RandomListNode> map = new HashMap<RandomListNode, RandomListNode>();        RandomListNode ret = new RandomListNode(head.label);        map.put(head, ret);        RandomListNode rhead = ret;        RandomListNode phead = head;        while (head != null) {            RandomListNode node = new RandomListNode(head.label);            map.put(head, node);            ret.next = node;            ret = ret.next;            head = head.next;        }        ret = rhead;        head = phead;        while(head != null) {            ret.random = map.get(head.random);            ret = ret.next;            head = head.next;        }        return rhead;      }

/*HashMap的key存原始pointer，value存新的pointer。第一遍，先不copy random的值，只copy數值建立好新的鏈表。並把新舊pointer存在HashMap中。第二遍，遍曆舊錶，複製random的值，因為第一遍已經把鏈表複製好了並且也存在HashMap裡了，所以只需從HashMap中，把當前舊的node.random作為key值，得到新的value的值，並把其賦給新node.random就好。 * */

    public RandomListNode copyRandomList1(RandomListNode head) {          if (head == null)              return head;          /*第一步:在OldList的每個節點後面都插入一個copyNode(拷貝鏈表的結點)*/          RandomListNode nowNode = head;          while (nowNode != null){              RandomListNode copyNode = new RandomListNode(nowNode.label);              copyNode.random = nowNode.random;              copyNode.next = nowNode.next;              nowNode.next = copyNode;              nowNode = nowNode.next.next;          }          /*第二步:確定NewList的每個節點,真正關聯到的Random結點是哪個，          *      因為第一步已經把所有NewList上的結點都建立了*/          nowNode = head;          while (nowNode != null){              if (nowNode.random != null){                  nowNode.next.random = nowNode.random.next;              }              nowNode = nowNode.next.next;          }          /*第三步：還原OldList的next為一開始的next結點          *      並拼接NewList的next到它真正所應該關聯的next結點          *      即：保持老鏈表OldList不變，拼接新鏈表NewList。          * */          RandomListNode pHead = new RandomListNode(0);          pHead.next = head;          RandomListNode newlist = pHead;          nowNode = head;          while (nowNode != null){              pHead.next = nowNode.next;              nowNode.next = pHead.next.next;              pHead = pHead.next;              nowNode = nowNode.next;          }          return newlist.next;      }