想把下面的字符串数字后的空格改为逗号？请问使用Java怎么实现？

java中正则表达式的匹配其实是利用不确定的有穷自动机（NFA）结合向上追溯的算法来实现的。以下是示例代码：import java.util.* ;/** An NFAState is a node with a set of outgoing edges to other* NFAStates.** There are two kinds of edges:** (1) Empty edges allow the NFA to transition to that state without* consuming a character of input.** (2) Character-labelled edges allow the NFA to transition to that* state only by consuming the character on the label.**/class NFAState{/** WARNING:** The maximum integer character code we'll match is 255, which* is sufficient for the ASCII character set.** If we were to use this on the Unicode character set, we'd get* an array index out-of-bounds exception.** A ``proper'' implementation of this would not use arrays but* rather a dynamic data structure like Vector.*/public static final int MAX_CHAR = 255 ;public boolean isFinal = false ;private ArrayList<NFAState> onChar[] = new ArrayList[MAX_CHAR] ;private ArrayList<NFAState> onEmpty = new ArrayList() ;/** Add a transition edge from this state to next which consumes* the character c.*/public void addCharEdge(char c, NFAState next) {onChar[(int)c].add(next) ;}/** Add a transition edge from this state to next that does not* consume a character.*/public void addEmptyEdge(NFAState next) {onEmpty.add(next) ;}public NFAState () {for (int i = 0; i < onChar.length; i++)onChar[i] = new ArrayList() ;}public boolean matches(String s) {return matches(s,new ArrayList()) ;}private boolean matches(String s, ArrayList visited) {/** When matching, we work character by character.** If we're out of characters in the string, we'll check to* see if this state if final, or if we can get to a final* state from here through empty edges.** If we're not out of characters, we'll try to consume a* character and then match what's left of the string.** If that fails, we'll ask if empty-edge neighbors can match* the entire string.** If that fails, the match fails.** Note: Because we could have a circular loop of empty* transitions, we'll have to keep track of the states we* visited through empty transitions so we don't end up* looping forever.*/if (visited.contains(this))/* We've found a path back to ourself through empty edges;* stop or we'll go into an infinite loop. */return false ;/* In case we make an empty transition, we need to add this* state to the visited list. */visited.add(this) ;if (s.length() == 0) {/* The string is empty, so we match this string only if* this state is a final state, or we can reach a final* state without consuming any input. */if (isFinal)return true ;/* Since this state is not final, we'll ask if any* neighboring states that we can reach on empty edges can* match the empty string. */for (NFAState next : onEmpty) {if (next.matches("",visited))return true ;}return false ;} else {/* In this case, the string is not empty, so we'll pull* the first character off and check to see if our* neighbors for that character can match the remainder of* the string. */int c = (int)s.charAt(0) ;for (NFAState next : onChar[c]) {if (next.matches(s.substring(1)))return true ;}/* It looks like we weren't able to match the string by* consuming a character, so we'll ask our* empty-transition neighbors if they can match the entire* string. */for (NFAState next : onEmpty) {if (next.matches(s,visited))return true ;}return false ;}}}/** Here, an NFA is represented by an entry state and an exit state.** Any NFA can be represented by an NFA with a single exit state by* creating a special exit state, and then adding empty transitions* from all final states to the special one.**/public class NFA{public NFAState entry ;public NFAState exit ;public NFA(NFAState entry, NFAState exit) {this.entry = entry ;this.exit = exit;}public boolean matches(String str) {return entry.matches(str);}/** c() : Creates an NFA which just matches the character `c'.*/public static final NFA c(char c) {NFAState entry = new NFAState() ;NFAState exit = new NFAState() ;exit.isFinal = true ;entry.addCharEdge(c,exit) ;return new NFA(entry,exit) ;}/** e() : Creates an NFA which matches the empty string.*/public static final NFA e() {NFAState entry = new NFAState() ;NFAState exit = new NFAState() ;entry.addEmptyEdge(exit) ;exit.isFinal = true ;return new NFA(entry,exit) ;}/** rep() : Creates an NFA which matches zero or more repetitions* of the given NFA.*/public static final NFA rep(NFA nfa) {nfa.exit.addEmptyEdge(nfa.entry) ;nfa.entry.addEmptyEdge(nfa.exit) ;return nfa ;}/** s() : Creates an NFA that matches a sequence of the two* provided NFAs.*/public static final NFA s(NFA first, NFA second) {first.exit.isFinal = false ;second.exit.isFinal = true ;first.exit.addEmptyEdge(second.entry) ;return new NFA(first.entry,second.exit) ;}/** or() : Creates an NFA that matches either provided NFA.*/public static final NFA or(NFA choice1, NFA choice2) {choice1.exit.isFinal = false ;choice2.exit.isFinal = false ;NFAState entry = new NFAState() ;NFAState exit = new NFAState() ;exit.isFinal = true ;entry.addEmptyEdge(choice1.entry) ;entry.addEmptyEdge(choice2.entry) ;choice1.exit.addEmptyEdge(exit) ;choice2.exit.addEmptyEdge(exit) ;return new NFA(entry,exit) ;}/* Syntactic sugar. */public static final NFA re(Object o) {if (o instanceof NFA)return (NFA)o ;else if (o instanceof Character)return c((Character)o) ;else if (o instanceof String)return fromString((String)o) ;else {throw new RuntimeException("bad regexp") ;}}public static final NFA or(Object... rexps) {NFA exp = rexps[0] ;for (int i = 1; i < rexps.length; i++) {exp = or(exp,re(rexps[i])) ;}return exp ;}public static final NFA s(Object... rexps) {NFA exp = e() ;for (int i = 0; i < rexps.length; i++) {exp = s(exp,re(rexps[i])) ;}return exp ;}public static final NFA fromString(String str) {if (str.length() == 0)return e() ;elsereturn s(re(str.charAt(0)),fromString(str.substring(1))) ;}public static void main(String[] args) {NFA pat = s(rep(or("foo","bar")),"") ;String[] strings ={ "foo" , "bar" ,"foobar", "farboo", "boofar" , "barfoo" ,"foofoobarfooX" ,"foofoobarfoo" ,} ;for (String s : strings) {System.out.println(s + "\t:\t" +pat.matches(s)) ;}}}

想把下面的字符串数字后的空格改为逗号？请问使用Java怎么实现？

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