001 /* 002 * Copyright (C) 2010 eXo Platform SAS. 003 * 004 * This is free software; you can redistribute it and/or modify it 005 * under the terms of the GNU Lesser General Public License as 006 * published by the Free Software Foundation; either version 2.1 of 007 * the License, or (at your option) any later version. 008 * 009 * This software is distributed in the hope that it will be useful, 010 * but WITHOUT ANY WARRANTY; without even the implied warranty of 011 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU 012 * Lesser General Public License for more details. 013 * 014 * You should have received a copy of the GNU Lesser General Public 015 * License along with this software; if not, write to the Free 016 * Software Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 017 * 02110-1301 USA, or see the FSF site: http://www.fsf.org. 018 */ 019 020 package org.crsh.util; 021 022 import java.util.ArrayList; 023 import java.util.Arrays; 024 import java.util.List; 025 import java.util.regex.Matcher; 026 import java.util.regex.Pattern; 027 028 /** 029 * @author <a href="mailto:julien.viet@exoplatform.com">Julien Viet</a> 030 * @version $Revision$ 031 */ 032 public class Strings { 033 034 /** . */ 035 private static final Pattern p = Pattern.compile("\\S+"); 036 037 public static List<String> chunks(CharSequence s) { 038 List<String> chunks = new ArrayList<String>(); 039 Matcher m = p.matcher(s); 040 while (m.find()) { 041 chunks.add(m.group()); 042 } 043 return chunks; 044 } 045 046 /** 047 * @see #findLongestCommonPrefix(Iterable) 048 */ 049 public static String findLongestCommonPrefix(CharSequence... seqs) { 050 return findLongestCommonPrefix(Arrays.asList(seqs)); 051 } 052 053 /** 054 * Find the longest possible common prefix of the provided char sequence. 055 * 056 * @param seqs the sequences 057 * @return the longest possible prefix 058 */ 059 public static String findLongestCommonPrefix(Iterable<? extends CharSequence> seqs) { 060 String common = ""; 061 out: 062 while (true) { 063 String candidate = null; 064 for (CharSequence s : seqs) { 065 if (common.length() + 1 > s.length()) { 066 break out; 067 } else { 068 if (candidate == null) { 069 candidate = s.subSequence(0, common.length() + 1).toString(); 070 } else if (s.subSequence(0, common.length() + 1).toString().equals(candidate)) { 071 // Ok it is a prefix 072 } else { 073 break out; 074 } 075 } 076 } 077 if (candidate == null) { 078 break; 079 } else { 080 common = candidate; 081 } 082 } 083 return common; 084 } 085 }