Finite state machine FST

Today, I read an article about lucene's use of finite state machines, http://www.cnblogs.com/LBSer/p/4119841.html, which at first felt very similar to trie trees, but later found that they were different:

Trie tree is a tree structure, each child node has only one parent node, while FST is a mesh structure, each child node can have multiple parent nodes, so FST saves more space.

I also saw an article that also mentioned the application of finite state machines.

Https://www.cnblogs.com/dreamroute/p/8484457.html

The use of finite State Machine in luence

The reason why String inputs= {"abc", "abd", "acf", "acg"}; / / keyslong outputs= {1rec 3je 5je 7}; / / valuesFST fst=new FST (); for (int item0 / sensitiveWordList.add (I, I + length) {sensitiveWordList.add (I, I + length)) {sensitiveWordList.add (I, I + length); / / minus 1 is because for increases itself by I = I + length-1. }} return sensitiveWordList;} / * replace sensitive word characters * * @ param txt * @ param matchType * @ param replaceChar * @ return * / public String replaceSensitiveWord (String txt, int matchType, String replaceChar) {String resultTxt = txt; / / get all sensitive words Set set = getSensitiveWord (txt, matchType) Iterator iterator = set.iterator (); String word = null; String replaceString = null; while (iterator.hasNext ()) {word = iterator.next (); replaceString = getReplaceChars (replaceChar, word.length ()); resultTxt = resultTxt.replaceAll (word, replaceString);} return resultTxt } / * get the replacement string * * @ param replaceChar * @ param length * @ return * / private String getReplaceChars (String replaceChar, int length) {String resultReplace = replaceChar; for (int I = 1; I < length; iTunes +) {resultReplace + = replaceChar;} return resultReplace } / * check whether the text contains sensitive characters. The check rules are as follows:

* if it exists, the length of sensitive word characters is returned. There is no return 0 * * @ param txt * @ param beginIndex * @ param matchType * @ return * / public int CheckSensitiveWord (String txt, int beginIndex, int matchType) {/ / sensitive word ending identification bit: boolean flag = false for sensitive words with only 1 digit / / the default number of matching identifiers is 0 int matchFlag = 0; Map nowMap = this.sensitiveWordMap; for (int I = beginIndex; I < txt.length (); iTunes +) {char word = txt.charAt (I); / / get the specified key nowMap = (Map) nowMap.get (word) / / if it exists, determine whether it is the last if (nowMap! = null) {/ / find the corresponding key and match the identity + 1 matchFlag++ / / if it is the last matching rule, end the loop and return the number of matching identifiers if ("1" .equals (nowMap.get ("isEnd") {/ / the ending flag bit is true flag = true / / the minimum rule is returned directly, and the maximum rule needs to continue searching for if (1 = = matchType) {break;} / / does not exist, and directly returns else {break }} / / length must be greater than or equal to 1, which is the word if (matchFlag < 2 | |! flag) {matchFlag = 0;} return matchFlag;} private Set readSensitiveWordFile () {Set wordSet = new HashSet (); wordSet.add ("× ×"); wordSet.add ("× ×"); wordSet.add ("son of a bitch") WordSet.add ("son of a bitch"); return wordSet;} public static void main (String [] args) {All all=new All (); all.sensitiveWordMap=all.addSensitiveWordToHashMap (all.readSensitiveWordFile ()); String txt = "× × ×"; String hou = all.replaceSensitiveWord (txt, 1, "*"); System.out.println ("text before replacement is:" + txt) System.out.println ("the replaced text is:" + hou);}}

I feel that this is very similar to the trie tree I wrote before. DFA seems to be a tree, not a mesh.