1.在一些字符串數組中，常會有重復的記錄，比如手機號碼，我們可以通過Hashtable來對其進行過濾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]))??????????????? hash.put(str[i], str[i]);??????? }??????? Enumeration enumeration=hash.keys();??????? 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]);?????? 輸出結果為：??????? 13811071503??????? 13811071501??????? 138110715002.A,B均為字符串數組，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]);輸出結果：??? 13811071503??? 13811071501存在的問題：每次都是倒序，可以再對程序稍加改動，變成正序。3.將一個字符串數組中某一個特定的字符串過濾掉/** *//**檢驗一個字符串數組，若包含某一特定的字符串，則將該字符串從數組中刪除，返回剩余的字符串數組???? * @param str_array 字符串數組???? * @param str_remove 待刪除的字符串???? * @return 過濾后的字符串???? */??? public String[] removeStrFromArray(String[] str_array,Stringstr_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]);??????? }??????? //生成一個新的數組??????? 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;??? }
posted on 2007-11-06 21:07
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