1.一个是对象里面的时间转时间戳

 2.然后用冒泡排序进行时间戳值的比较

 3.通过比较大小交换集合里面的对象进行排序

//获得列表结果查询
    @RequestMapping(value = "/feelist",method = RequestMethod.GET)
    @ResponseBody
    public ApiResponseResult feelist( long hotelId,String roomNo){
        Hotel h = hotelService.getHotel(hotelId);
        FeeMonth f=new FeeMonth();
        f.setHotel(h);
        f.setRoomNo(roomNo);
        List<FeeMonth> feelist =feeMonthService.FeeMonthgz(f);                      
        if(feelist==null || feelist.size()<1){
            return ApiResponseResult.failure("没有记录", -1);  
        }
        SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-MM-dd");

             for(int i=0;i<feelist.size()-1;i++){//外层循环控制排序趟数
            for(int j=0;j<feelist.size()-1-i;j++){//内层循环控制每一趟排序多少次
            try {
                if((simpleDateFormat.parse(feelist.get(j).getDatemonth())).getTime()>(simpleDateFormat.parse(feelist.get(j+1).getDatemonth())).getTime()){
                FeeMonth fee=feelist.get(j);
                feelist.set((j), feelist.get(j+1));  
                feelist.set((j+1), fee);  
                }
            } catch (ParseException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            }
            } 

        return ApiResponseResult.success().data(feelist);
    }//..