【Java】对ArrayList排序

java如何对ArrayList中对象按照该对象某属性排序

（从小到大）

两种方法：

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方法一：<br>Comparator<KNNNode> comparator = new Comparator<KNNNode>() {           @Override         public int compare(KNNNode o1, KNNNode o2) {             System.out.println("进入 compare function");             // TODO Auto-generated method stub             if (o1.getDistance() <= o2.getDistance()) {//o1属性小于等于o2属性时，返回1                 return 1;             } else {                 return -1;             }         }       };

 

调用方法：

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List<KNNNode> nodelist = new ArrayList<KNNNode>(); //此处需要填充nodelist.add(); Collections.sort(nodelist, comparator);// 调用

 

 

 

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方法二：<br>class SortBy implements Comparator {         public int compare(Object obj1, Object obj2) {             KNNNode point1 = (KNNNode) obj1;             KNNNode point2 = (KNNNode) obj2;             if (point1.getDistance() <= point2.getDistance()) {                 return 1;             } else {                 return 0;             }         }     }

 

调用方法：

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1	Collections.sort(nodelist, new SortBy());

KNNNode结点内的属性：

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package newKNN;   /**  * KNN结点类，用来存储最近邻的k个元组相关的信息  */ public class KNNNode {     private int index; // 元组标号     private double distance; // 与测试元组的距离     private String seq;//滑动窗口motif序列片段           public KNNNode(int index, double distance, String seq) {         super();         this.index = index;         this.distance = distance;         this.seq = seq;               }     public int getIndex() {         return index;     }     public void setIndex(int index) {         this.index = index;     }     public double getDistance() {         return distance;     }     public void setDistance(double distance) {         this.distance = distance;     }     public String getSeq() {         return seq;     }     public void setSeq(String seq) {         this.seq = seq;     }       }

 我想要按照KNNNode的distance属性排序。

 

 

若要按照从大到小的顺序排序，如果只是将

>=返回值改为-1

<返回值改为1 会出现异常。

Comparison method violates its general contract!

 原因是是实现接口Comparator 的类有问题，在compare中没有处理好两个比较的对象相等的情况

public int compare(KNNNode o1, KNNNode o2）;

即 o1.getDistance() == o2.getDistance() 时，返回值不是0

我最终的实现代码是：（从大到小排序）

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Comparator<KNNNode> comparator = new Comparator<KNNNode>() {           @Override         public int compare(KNNNode o1, KNNNode o2) {             // TODO Auto-generated method stub             if (o1.getDistance() < o2.getDistance()) {                 return 1;             } else if(o1.getDistance() > o2.getDistance()){                 return -1;             }else{                 return 0;             }         }       };

 

参考列表：

http://blog.csdn.net/telenewbie/article/details/45061089

http://www.cnblogs.com/xujian2014/p/5215082.html

http://www.cnblogs.com/fzzl/archive/2010/08/14/1799408.html