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這個程序應該是什麼樣轉換:如何檢查是否有新線?
\hyp76{a,1+a-b}b{xy}
到:
\HyperpFq{7}{6}@@{a,1+a-b}{b}{xy}.
正如你可以看到,如果一個組的內容是一個字符,則沒有括號。但是,當一個帶括號的組位於行尾時,它將跳過該組,並將其視爲單個字符(不帶括號)的情況。我怎樣才能避免這種情況?謝謝。這裏是我的代碼:
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();
非常混亂。嘗試澄清你的問題。什麼是單個字符的情況?什麼是括號內的情況?什麼是返回0的無用函數'checkNestedBrackFront'? – djb