list排序的问题

1.需求，找出list对象中某个属性最大的
2.我的实现

 private Double getMaxUpperrangeLimit()
    {
        ParametersPoolInterface parameterPool = RequestContext.getPool();
        ListSensorParameter listSensorLp = (ListSensorParameter)parameterPool.getParameter(ParameterNames.LIST_SENSOR_LP);
        List<Sensor> sensorLpList = listSensorLp.getSensorList();
        if(sensorLpList == null || sensorLpList.size() < 1)
        {
            return null;
        }
        Collections.sort(sensorLpList, new Comparator<Sensor>()
                {
                    public int compare(Sensor o1, Sensor o2)
                    {
                        double v = o1.getUpperrangeLimit() - o2.getUpperrangeLimit();
                        if (v < 0)
                        {
                            return 1;
                        } else if (v > 0)
                        {
                            return -1;
                        } else
                        {
                            return 0;
                        }
                    }
                });
        return sensorLpList.get(0).getUpperrangeLimit();
    }

3.manager的实现

    private Double getMaxUpperrangeLimit()
    {
        ParametersPoolInterface parameterPool = RequestContext.getPool();
        ListSensorParameter listSensorLp = (ListSensorParameter)parameterPool.getParameter(ParameterNames.LIST_SENSOR_LP);
        List<Sensor> sensorLpList = listSensorLp.getSensorList();
        if (sensorLpList == null || sensorLpList.size() < 1)
        {
            return null;
        }
        Double max = sensorLpList.get(0).getUpperrangeLimit();
        for (int i = 1; i < sensorLpList.size(); i++ )
        {
            Double tmp = sensorLpList.get(i).getUpperrangeLimit();
            if (tmp > max)
            {
                max = tmp;
            }
        }
        return max;
    }

总结：我的写法缺点是把list原来的顺序给改变了，这就涉及到面试中常用的一个情况，不需要改变值的时候，可以讲这个list赋值给给另外一个list，然后再对另外一个list进行操作，算是用一个临时变量去操作！而且取一个最大值，只要遍历一次就可以了。