在java List中进行模糊查询的实现方法

比如我有下面这样一个list，里面存放的是多个employee对象。然后我想对这个list进行按照employee对象的名字进行模糊查询。有什么好的解决方案么？

比如我输入的查询条件为“wang”，那么应该返回只包含employee1的list列表。

list list = new arraylist();
employee employee1 = new employee();
employee1.setname("wangqiang");
employee1.setage(30);
list.add(employee1);
employee employee2 = new employee();
employee2.setname("lisi");
list.add(employee2);
employee2.setage(25);

方式一：

public list search(string name,list list){
  list results = new arraylist();
  pattern pattern = pattern.compile(name);
  for(int i=0; i < list.size(); i++){
   matcher matcher = pattern.matcher(((employee)list.get(i)).getname());
   if(matcher.matches()){
     results.add(list.get(i));
   }
  }
  return results;
}

上面那个是大小写敏感的，如果要求大小写不敏感，改成：

pattern pattern = pattern.compile(name,pattern.case_insensitive);

并且上面那个是精确查询，如果要模糊匹配，matcher.find()即可以进行模糊匹配

public list search(string name,list list){
  list results = new arraylist();
  pattern pattern = pattern.compile(name);
  for(int i=0; i < list.size(); i++){
   matcher matcher = pattern.matcher(((employee)list.get(i)).getname());
   if(matcher.find()){
     results.add(list.get(i));
   }
  }
  return results;
}

方式二：

public class listlike {
 
//定义员工类
public class employee {
private string name;
private int age;
public int getage() {
return age;
}
public void setage(int age) {
this.age = age;
}
public string getname() {
return name;
}
public void setname(string name) {
this.name = name;
}
}
 
public list list=new arraylist();
 
//增加员工
public list addlist(string name,int age){
employee employee1 = new employee();
employee1.setname(name);
employee1.setage(age);
list.add(employee1);
return list;
}
 
//显示所有员工
public void showlist(){
for(int i=0;i<list.size();i++){
system.out.println(((employee)(list.get(i))).getname()+" "+((employee)(list.get(i))).getage());
}
}
 
//模糊查询
public list likestring(string likename){
for(int i=0;i<list.size();i++){
if(((employee)(list.get(i))).getname().indexof(likename)<=-1)
list.remove(i);
}
return list;
 
}
 
public static void main(string arg[]){
listlike ll=new listlike();
ll.addlist("wuxiao",13);
ll.addlist("wangwang",11);
ll.addlist("wanghua",12);
ll.addlist("xiaowang",13);
ll.addlist("xiaoxiao",13);
 
ll.likestring("wang");
ll.showlist();
 
 
}
 
}