`
ihuashao
  • 浏览: 4533044 次
  • 性别: Icon_minigender_1
  • 来自: 济南
社区版块
存档分类
最新评论

使用Hashtable对字符串进行碰撞

阅读更多

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

publicString[]checkArray(String[]str)...{
Hashtable
<String,String>hash=newHashtable<String,String>();

for(inti=0;i<str.length;i++)...{
if(!hash.containsKey(str[i]))
hash.put(str[i],str[i]);
}


Enumerationenumeration
=hash.keys();
String[]str_new
=newString[hash.size()];
inti=0;

while(enumeration.hasMoreElements())...{
str_new[i]
=enumeration.nextElement().toString();
i
++;
}

returnstr_new;
}

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

/***//**检验一个字符串数组,若包含某一特定的字符串,则将该字符串从数组中删
除,返回剩余的字符串数组
*
@paramstr_array字符串数组
*
@paramstr_remove待删除的字符串
*
@return过滤后的字符串
*/

publicString[]removeStrFromArray(String[]str_array,String
str_remove)
...{
Hashtable
<String,String>hash=newHashtable<String,String>();
for(inti=0;i<str_array.length;i++)...{
if(!str_array[i].equals(str_remove))
hash.put(str_array[i],str_array[i]);
}

//生成一个新的数组
String[]str_new=newString[hash.size()];
inti=0;
Enumerationenumeration
=hash.keys();
while(enumeration.hasMoreElements())...{
str_new[i]
=enumeration.nextElement().toString();
i
++;
}

returnstr_new;
}

分享到:
评论

相关推荐

Global site tag (gtag.js) - Google Analytics