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这个程序应该是什么样转换:如何检查是否有新线?
\hyp76{a,1+a-b}b{xy}
到:
\HyperpFq{7}{6}@@{a,1+a-b}{b}{xy}.
正如你可以看到,如果一个组的内容是一个字符,则没有括号。但是,当一个带括号的组位于行尾时,它将跳过该组,并将其视为单个字符(不带括号)的情况。我怎样才能避免这种情况?谢谢。这里是我的代码:
static int checkNestedBrackFront(String line, int pos){
int count=0;
for(int i=pos;i<line.length();i++){
if(line.charAt(i)=='{')
count++;
if(line.charAt(i)=='}'&&count==0)
return i;
if(line.charAt(i)=='{')
count--;
}
return 0;
}
line = new Scanner(new File("KLSadd.tex")).useDelimiter("\\Z").next();
PrintWriter writer = new PrintWriter("Converted.tex");
while(line.contains("\\hyp76")){
int posHyp = line.indexOf("\\hyp76");
String beforeHyp = line.substring(0,posHyp);
int start = posHyp+7;
String firstArgs = line.substring(start, line.indexOf("}", start));
if(line.charAt((line.indexOf("}", start)+1))!='{'&&!(line.substring(start, start+4).contains("\n"))){ //this is to check for single characters
secArgs = line.substring(line.indexOf("}", start) + 1,line.indexOf("}", start) + 2);
posSec=line.indexOf("}", start)+1;
}
else {
int posBrack = line.indexOf("{", line.indexOf("}", start));
posSec = checkNestedBrackFront(line, posBrack+1);
secArgs = line.substring(posBrack+1, posSec);
}
if((line.charAt(posSec+1)!='{')&&!(line.substring(posSec,posSec+4).contains("\n"))){ //this is to check for single characters
System.out.println(line.charAt(posSec+1)+"hello");
thirdArgs= line.substring(posSec+1,posSec+2);
afterHyp=line.substring(posSec+2);
}
else{
int posThirdBrack = line.indexOf("{", posSec);
int posThird = checkNestedBrackFront(line, posThirdBrack+1);
thirdArgs = line.substring(posThirdBrack+1,posThird);
afterHyp = line.substring(posThird+1);
}
line=beforeHyp+"\\HyperpFq{7}{6}@@{"+firstArgs+"}{"+secArgs+"}{"+thirdArgs+"}"+afterHyp;
}
writer.print(line);
writer.close();
非常混乱。尝试澄清你的问题。什么是单个字符的情况?什么是括号内的情况?什么是返回0的无用函数'checkNestedBrackFront'? – djb