使用Hashtable对字符串进行碰撞-ZT

Posted on 2007-07-30 16:36 my 阅读(1044) 评论(0)  编辑  收藏 所属分类: 个人收藏
使用Hashtable对字符串进行碰撞

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

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

for(int i=0;i<str.length;i++)...{
if(!hash.containsKey(str[i])) //测试指定对象是否为此哈希表中的键
hash.put(str[i], str[i]);
}

Enumeration enumeration=hash.keys(); //返回此哈希表中的键的枚举
String[] str_new=new String[hash.size()];
int i=0;

while(enumeration.hasMoreElements())...{
str_new[i]=enumeration.nextElement().toString();
i++;
}
return str_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
* @param str_array 字符串数组
* @param str_remove 待删除的字符串
* @return 过滤后的字符串
*/
public String[] removeStrFromArray(String[] str_array,String
str_remove)...{
Hashtable<String, String> hash=new Hashtable<String, String>();
for(int i=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=new String[hash.size()];
int i=0;
Enumeration enumeration=hash.keys();
while(enumeration.hasMoreElements())...{
str_new[i]=enumeration.nextElement().toString();
i++;
}
return str_new;
}

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