List a = new ArrayList();
        Map b = new HashMap();
        b.put(1, 1);
        b.put(2,b);
        

        List list = (List)b.get(2);//取得集合a
        for(Object o : list){//循环集合a
            
        }

------解决方案--------------------

就是循环嵌套

服务器端处理
Java code

List<xxx> serverList = ...; //假设系统商品库集合,xxx是商品类

//接收处理
//假设用Object序列化传输
new ObjectInputStream ois = new ObjectInputStream(socket.getInputStream());
Map<Integer, Object> map = (Map<Integer, Object>)ois.readObject();
List<xxx> clientList = (List<xxx>)map.get(2);
Map<Integer, xxx> result = new HashMap<Integer, xxx>(); //返回给客户端
for (xxx a : clientList) {
    for (xxx b : serverList) {
        if (a.getId() == b.getId()) {//客户端和系统商品库集合id相同
            b.setAmount(b.getAmout() + a.getAmount()); //系统商品集合数量修改
              //....其他修改
        }
    }
    if (!result.containsKey(a.getId)) {
        result.put(a.getId(), new xxx());
    }
    xxx b = result.get(a.getId());
    b.setAmout(b.getAmout() + a.getAmout()); //返回给客户端的清单合计
    //...其他合计
}

------解决方案--------------------
就是个循环嵌套，有什么不能遍历的，赞成5楼！