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    /*******************************************************************************
    ?*
    ?* 對數字或文件名排列
    ?*
    ?* 沒有考慮漢字顯示時占兩個位置的情況,即對文件名處理不太好
    ?*
    ?* Author: NeedJava
    ?*
    ?* Modified: 2007.09.09
    ?*
    ?******************************************************************************/

    public final class DotAlign
    {
    ? private DotAlign(){}

    ? private static final String align( String string,
    ?????????????????????????????????????char prefix,
    ???????????????????????????????????? char suffix,
    ?????????????????????????????????????int prefixLength,
    ???????????????????????????????????? int suffixLength,
    ?????????????????????????????????????boolean isNumber )
    ? {
    ??? if( string==null||string.equals( "" )||string.equals( "." )||prefixLength<0||suffixLength<0 )
    ????? {
    ??????? return string;
    ????? }

    ??? //如果沒有小數,那么按照常理,小數點也應該省略
    ??? char[] buffer=new char[( suffixLength>0 ? prefixLength+1+suffixLength : prefixLength )];

    ??? int stringLength=string.length();

    ??? int dot=stringLength-1;

    ??? //找到小數點位置,可以用lastIndexOf()代替
    ??? while( dot>-1 )
    ???????? {
    ?????????? if( string.charAt( dot-- )=='.' )
    ???????????? {
    ?????????????? dot++;

    ?????????????? break;
    ???????????? }
    ???????? }

    ??? //System.out.print( Integer.toString( dot )+"\t" );

    ??? //★整數部分,從小數點左位向左移
    ??? int pointer=( dot>-1 ? dot-1 : stringLength-1 );

    ??? for( int i=prefixLength-1; i>-1; i-- )
    ?????? {
    ???????? if( pointer>-1 )
    ?????????? {
    ???????????? buffer[i]=string.charAt( pointer-- );
    ?????????? }
    ???????? else{ buffer[i]=prefix;
    ???????????? }
    ?????? }

    ??? //如果沒有小數部分,那么按照常理,小數點應該省略
    ??? if( suffixLength>0 )
    ????? {
    ??????? //★小數點
    ??????? buffer[prefixLength]=( dot>-1||isNumber ? '.' : suffix );

    ??????? //★小數部分,從小數點右位向右移
    ??????? pointer=( dot>-1 ? dot+1 : stringLength );

    ??????? for( int i=prefixLength+1; i<buffer.length; i++ )
    ?????????? {
    ???????????? if( pointer<stringLength )
    ?????????????? {
    ???????????????? buffer[i]=string.charAt( pointer++ );
    ?????????????? }
    ???????????? else{ buffer[i]=suffix;
    ???????????????? }
    ?????????? }
    ????? }

    ??? return new String( buffer );
    ? }


    ? public final static void main( String[] args )
    ? {
    ??? System.out.println( "\r\n-------------------------------------------\r\n" );

    ??? String[] number={ "",
    ????????????????????? ".",
    ????????????????????? ".5",
    ????????????????????? ".56",
    ????????????????????? ".567",
    ????????????????????? ".5678",
    ????????????????????? "1234",
    ????????????????????? "1234.",
    ????????????????????? "1234.5",
    ????????????????????? "1234.56",
    ????????????????????? "1234.567",
    ????????????????????? "1234.5678" };

    ??? for( int i=0; i<number.length; i++ )
    ?????? {
    ???????? System.out.println( "["+DotAlign.align( number[i], ' ', '0', 7, 2, true )+"]?? ["+number[i]+"]" );
    ?????? }

    ??? System.out.println( "\r\n-------------------------------------------\r\n" );

    ??? String[] file={ "",
    ??????????????????? ".",
    ??????????????????? ".j",
    ??????????????????? ".jp",
    ??????????????????? ".jpe",
    ??????????????????? ".jpeg",
    ??????????????????? "1234",
    ??????????????????? "1234.",
    ??????????????????? "1234.j",
    ??????????????????? "1234.jp",
    ??????????????????? "1234.jpe",
    ??????????????????? "1234.jpeg" };

    ??? for( int i=0; i<file.length; i++ )
    ?????? {
    ???????? System.out.println( "["+DotAlign.align( file[i], ' ', ' ', 7, 5, false )+"]?? ["+file[i]+"]" );
    ?????? }

    ??? System.out.println( "\r\n-------------------------------------------\r\n" );
    ? }
    }

    posted on 2007-09-14 00:17 NeedJava 閱讀(1132) 評論(0)  編輯  收藏 所屬分類: Java
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