1.在一些字符串数组中，常会有重复的记录，比如手机号码，我们可以通过Hashtable来对其进行过滤
示例：
String[] mobile={"13811071500","13811071500","13811071501","13811071503","13811071501"};
mobile=checkArray(mobile);
for(int i=0;i<mobile.length;i++)
System.out.println(mobile[i]);
输出结果为：
13811071503
13811071501
13811071500
2.A,B均为字符串数组，找出在A中存在，而在B中不存在的字符串
public String[] compareArray(String[] A,String[] B){
Hashtable<String, String> hash=new Hashtable<String, String>();
Hashtable<String, String> hash_new=new Hashtable<String, String>();
for(int i=0;i<B.length;i++)
hash.put(B[i], B[i]);
for(int i=0;i<A.length;i++){
if(!hash.containsKey(A[i]))
hash_new.put(A[i], A[i]);
}
String[] C=new String[hash_new.size()];
int i=0;
Enumeration enumeration=hash_new.keys();
while(enumeration.hasMoreElements()){
C[i]=enumeration.nextElement().toString();
i++;
}
return C;
}
示例：
String[] mobile1={"13811071500","13811071501","13811071502","13811071503","13811071504"};
String[] mobile2={"13811071500","13811071505","13811071502","13811071506","13811071504"};
String[] mobile3=compareArray(mobile1,mobile2);
for(int i=0;i<mobile3.length;i++)
System.out.println(mobile[i]);
输出结果：
13811071503
13811071501
存在的问题：
每次都是倒序，可以再对程序稍加改动，变成正序。
3.将一个字符串数组中某一个特定的字符串过滤掉
