利用java做排序

一、利用TreeMap做排序，代码如下：

java.util.SortedMap<Double, scala.Tuple2<String,String>> topN = new java.util.TreeMap<Double, scala.Tuple2<String,String>>();


	    topN.put(-0.223,new scala.Tuple2<String,String>("1","1"));
	    topN.put(-0.5239,new scala.Tuple2<String,String>("2","1"));
	    topN.put(-0.323,new scala.Tuple2<String,String>("3","1"));
	    topN.put(-0.423,new scala.Tuple2<String,String>("4","1"));
	    topN.put(-0.523,new scala.Tuple2<String,String>("5","1"));
	    topN.put(-0.623,new scala.Tuple2<String,String>("6","1"));
	     
	    for (Map.Entry<Double, scala.Tuple2<String,String>> entry : topN.entrySet()) {
	    	 System.out.println(entry.getValue()._1); 
	     }

结果发现，排序输出没有问题，都是从小到大输出；但是如果存在排序值相同的数据时，就会出现数据覆盖。
二、利用Collections.sort 做排序，则不会出现map中的覆盖问题，代码如下：

 List<Tuple2<Double,Tuple2<String,String>>> listA = new ArrayList<Tuple2<Double,Tuple2<String,String>>>();
	    
	    listA.add(new Tuple2<Double,Tuple2<String,String>>(-0.223,new scala.Tuple2<String,String>("1","1")));
	    listA.add(new Tuple2<Double,Tuple2<String,String>>(-0.5239,new scala.Tuple2<String,String>("2","1")));
	    listA.add(new Tuple2<Double,Tuple2<String,String>>(-0.323,new scala.Tuple2<String,String>("3","1")));
	    listA.add(new Tuple2<Double,Tuple2<String,String>>(-0.423,new scala.Tuple2<String,String>("4","1")));
	    listA.add(new Tuple2<Double,Tuple2<String,String>>(-0.523,new scala.Tuple2<String,String>("5","1")));
	    listA.add(new Tuple2<Double,Tuple2<String,String>>(-0.623,new scala.Tuple2<String,String>("6","1")));
	    
	    Collections.sort(listA, new Comparator<Tuple2<Double,Tuple2<String,String>>>() {
            public int compare(Tuple2<Double,Tuple2<String,String>> arg0, Tuple2<Double,Tuple2<String,String>> arg1) {
            	Double d0=arg0._1;
            	Double d1=arg1._1;
            	return d0.compareTo(d1); 
            }
        });
         
        for (Tuple2<Double,Tuple2<String,String>> p : listA) {
            System.out.println(p._2._1);
        }