Java-生成一个36位的GUID.doc

publicstaticvoidmain(String args) / 8 4 4 4 12/ 44CB8C3B-1CBB-3BDA-1B03-F3E18D42C59D/ 6C024CC1-99E3-F88B-2C57-F0907A7A1801/ E337CD73-4814-15A6-AB20-9C73ABDE7875/ E823A410-9B83-E12E-01DA-50C99CFFB96B/ 1B8A5253-BC6D-C977-F9CD-31A01A1FCF9EcharA_Z = A,B,C,D,E,F;List guidList=newArrayList();for(intj=0;j50000;j+)String guid =;String syw=;/上一位for(inti = 0; i 36; i+) if(i = 8 | i = 13 | i = 18 | i = 23) guid = guid +-;elseinttt = (int) (Math.random() * 1000);inttt1 = (int) (Math.random() * 100);intnum = tt - tt1;if(num 0 & !syw.equals(0) guid = guid +0;syw=0;elseString str = num +;intlastNum = Integer.parseInt(str.substring(str.length() - 1);if(lastNum A_Z.length) /为避免在新生成的guid中，数字只有数字6以上的，下面做了处理if(tt-i)%2=0 & !syw.equals(lastNum+)guid = guid + lastNum;syw=lastNum+;elseif(!syw.equalsIgnoreCase(A_ZlastNum+)guid = guid + A_ZlastNum;syw=A_ZlastNum+;elseif(lastNum=0)guid = guid + A_ZlastNum+1;syw=A_ZlastNum+1+;elseguid = guid + A_ZlastNum-1;syw=A_ZlastNum-1+;else/为避免6,7,8,9 出现多次，下面做了处理if(tt-lastNum)%2=0 & !syw.equals(10-lastNum)+)guid = guid + (10-lastNum);syw=(10-lastNum)+;elseif(!syw.equals(lastNum+)guid = guid + lastNum;syw=lastNum+;elseguid = guid + (lastNum-1);syw=(lastNum-1)+;guidList.add(guid);System.out.println(guidList.size();for(intg=0;gguidList.size();g+)String guid=guidList.get(g).trim();/0/判断当前的guid中相邻的两个字符是否相同/charguidch=guid.toCharArray();/for(into=0;oguidch.length-2;o+)/if(guidcho=guidcho+1)/System.out.println(guid);/break;/判断当前X万条guid中是否有重复的for(intu=0;uguidList.size();u+)if(g!=u)String guids=guidL