面试碰到几回list的去重和排序。下面介绍一种作法:html
List<Student>容量10k以上,要求去重复。这里Student的重复标准是属性相同,所以须要重写equals和hashcode方法,不知道有几个能够手写出来。java
student的equals方法:面试
public void equals(Object o){ if(this == o) retun true; if(!(o instanceof Student)) return false; Student stu = (Studend)o; if(id!=stu.id) return false; if(age!=stu.age) return false; return name!=null ? name.equals(stu.name) : stu.name ==null; }
这里只要记住宗旨是比较Student的属性便可,若是属性相同则相等。先考虑地址相等,而后类型匹配instanceof。接下来是各类属性,int属性直接双等号比较,String类型须要判断是否为null,若是是null则都是null返回true,若是不是null则比较equals。算法
student的hashcode方法:数组
public int hashCode(){ int result = id; reuslt = 31*id +(name!=null?name.hashCode():0); reuslt = 31*age; return reuslt; }
hashCode是为了hash表计算作辅助,方便快速查找。所以hash算法的结果要尽可能的散列。这里用到31,这个31在别的博客中看到的缘由是这样的:obj*31==obj<<5-obj.左移5位至关乘以2的5次方,就是32.null的hashCode为空。dom
经过equals和hashCode的实现能够发现,若是equals为true,则全部属性相同,而属性相同则计算出的hashCode必然相同。然而hashCode相同,属性未必同样,即equals不必定为真。ide
关于hashCode的价值体现并不在这里,而在于HashMap的实现。HashMap内部是经过链表数组的hash结构来实现的,这里就要用到hashcode。this
下面是完整的Student代码:spa
package com.test.arithmetic.listequals; /** * 这里id,name,age相同则Student相同, * 如有其余相同 * Created by Administrator on 2016/3/29. */ public class Student { int id; String name; int age; public Student(int id, String name, int age) { this.id = id; this.name = name; this.age = age; } @Override public boolean equals(Object o) { if (this == o) return true; if (!(o instanceof Student)) return false; Student student = (Student) o; if (id != student.id) return false; if (age != student.age) return false; return name != null ? name.equals(student.name) : student.name == null; } @Override public int hashCode() { int result = id; result = 31 * result + (name != null ? name.hashCode() : 0); result = 31 * result + age; return result; } }
若是你以为本身能够hold住一个完善的hash算法就能够本身去实现它。这里采用jdk自带的HashSet来完成重复获取。3d
先放代码:
package com.test.arithmetic.listequals; import org.junit.Assert; import java.util.*; /** * 取出list中重复的Student对象 * Created by Administrator on 2016/3/29. */ public class ObtainListEquals { public static void main(String[] args){ //原始数据 List<Student> list = new ArrayList<>(); //重复数据 List<Student> list2 = new ArrayList<>(); //填充 for (int i = 0; i < 10 ; i++) { list.add(new Student(i,"_"+i,18+i)); Random random = new Random(); if (random.nextBoolean()){ list.add(new Student(i,"_"+i,18+i)); } } //使用hashset去重复,set为重复的集合,能够经过new ArrayList(set)转换成list HashSet<Student> set = new HashSet<>(); for (Student student : list) { boolean add = set.add(student); if (!add){ list2.add(student); } } //比较 Assert.assertEquals(list.size(),list2.size()+set.size()); } }
去重的原理和简单,不管你仅仅是想把重复的丢掉,或者将重复的取出来。这里去掉的是第二次遇到的对象,取出的也是第二次遇到的对象。HashSet中的add方法会返回一个Boolean值,若是插入的值已经存在,则直接返回false。关于hashset的源码放到之后研究。大概的说,是经过HashMap的key来实现的,而HashMap在1.8中改动很大,听说是用红黑树实现的,提升了get的时间复杂度。参考:1.8HashMap
一样list中存放的是Student对象,我须要一个规则来排序。这个排序的规则这里定义为id的比较大小。参考:java中list排序
Comparable接口提供一个比较的compareTo(Object o)方法,经过返回值>0,=0,<0比较大小。这里因为仅仅把id当作比较大小的方法,直接用id作减法,若是是要比较对象,建议套用this.property.compareTo(o.property).
package com.test.arithmetic.listequals; /** * 这里id,name,age相同则Student相同, * 如有其余相同 * Created by Administrator on 2016/3/29. */ public class Student implements Comparable<Student>{ int id; String name; int age; public Student(int id, String name, int age) { this.id = id; this.name = name; this.age = age; } @Override public boolean equals(Object o) { if (this == o) return true; if (!(o instanceof Student)) return false; Student student = (Student) o; if (id != student.id) return false; if (age != student.age) return false; return name != null ? name.equals(student.name) : student.name == null; } @Override public int hashCode() { int result = id; result = 31 * result + (name != null ? name.hashCode() : 0); result = 31 * result + age; return result; } @Override public int compareTo(Student o) { return this.id-o.id; } }
经过Collections.sort(list)排序:
package com.test.arithmetic.list.sort; import com.test.arithmetic.list.Student; import org.junit.Before; import org.junit.Test; import java.util.ArrayList; import java.util.Collections; import java.util.List; /** * 对list中对象排序 * Created by Administrator on 2016/3/29. */ public class SortList { List<Student> list; @Before public void setUp(){ list = new ArrayList<>(); for (int i = 0; i < 10; i++) { int v = (int)(Math.random() * 100); list.add(new Student(v,"_"+v,18+v)); } System.out.println("原list:"+list); } //方法一,对象实现Comparable接口 @Test public void byImplements(){ Collections.sort(list); System.out.println("排序后:"+list); } }
Student类仍是未实现Comparable接口以前的:
package com.test.arithmetic.list; /** * 这里id,name,age相同则Student相同, * 如有其余相同 * Created by Administrator on 2016/3/29. */ public class Student{ int id; String name; int age; public Student(int id, String name, int age) { this.id = id; this.name = name; this.age = age; } public int getId() { return id; } public Student(int id) { this.id = id; } @Override public boolean equals(Object o) { if (this == o) return true; if (!(o instanceof Student)) return false; Student student = (Student) o; if (id != student.id) return false; if (age != student.age) return false; return name != null ? name.equals(student.name) : student.name == null; } @Override public int hashCode() { int result = id; result = 31 * result + (name != null ? name.hashCode() : 0); result = 31 * result + age; return result; } @Override public String toString() { return "Student{" + "id=" + id + ", name='" + name + '\'' + ", age=" + age + '}'; } }
在排序的代码出添加排序规则:
package com.test.arithmetic.list.sort; import com.test.arithmetic.list.Student; import org.junit.Before; import org.junit.Test; import java.util.ArrayList; import java.util.Collections; import java.util.Comparator; import java.util.List; /** * 对list中对象排序 * Created by Administrator on 2016/3/29. */ public class SortList { List<Student> list; @Before public void setUp(){ list = new ArrayList<>(); for (int i = 0; i < 10; i++) { int v = (int)(Math.random() * 100); list.add(new Student(v,"_"+v,18+v)); } System.out.println("原list:"+list); } //方法一,对象实现Comparable接口 @Test public void byImplements(){ // Collections.sort(list); System.out.println("排序后:"+list); } /*方法二,添加比较器*/ @Test public void byOverideCompare(){ Collections.sort(list, new Comparator<Student>() { @Override public int compare(Student o1, Student o2) { return o1.getId()-o2.getId(); } }); System.out.println(list); } }