Java 匹配包含完整字符串分隔符的子字符串
我不知道该如何表达这个问题。长话短说,我想从a(b)中的Java 匹配包含完整字符串分隔符的子字符串,java,regex,pattern-matching,match,Java,Regex,Pattern Matching,Match,我不知道该如何表达这个问题。长话短说,我想从a(b)中的行中拉出两个字符串(a,b)。几乎在所有情况下,a=b,但为了以防万一,我把它们分开了。问题:两个字符串都可以包含任何字符,包括Unicode、空格、标点和括号 1: In: ThisName (ThisName) is in this list 2: In: OtherName (With These) (OtherName (With These)) is in this list 3: In: Really Annoying (Bec
行中拉出两个字符串(a
,b
)。几乎在所有情况下,a=b,但为了以防万一,我把它们分开了。问题:两个字符串都可以包含任何字符,包括Unicode、空格、标点和括号
1: In: ThisName (ThisName) is in this list
2: In: OtherName (With These) (OtherName (With These)) is in this list
3: In: Really Annoying (Because) Separators (Really Annoying (Because) Separators) is in this list
第1行,简单:^\w+:\s(?'a'.+?)\s\((?'b'.+)\)
a:ThisName
b:ThisName
第2行,与前面相同:a:OtherName
b:With this)(OtherName(With this)
第2行,lazy:^\w+:\s(?'a'.+?)\s\((?'b'.+?)\)
a:OtherName
b:With the this
3号线,总台
这可能吗?也许我需要走另一条路线?我们知道需要一组括号。也许我必须走一条数学路线,计算括号的数量,找到那条路线来确定哪个应该包含b
?以某种方式计算每个打开和关闭
我一直在玩的东西:
顺便说一下,如果我能改变输入格式,我肯定会的
补充问题:如果这不可能,那么一直假设
a=b
会让这更容易吗?我想不出会是什么样子。我会做的是不使用正则表达式。遵循这种算法:
看起来可能有多个字符串组成“B”(从第3行开始),因此您可以按照上面的步骤2不断迭代字符串,将字符串添加到列表或字符串生成器中(视情况而定)。我要做的是不使用正则表达式。遵循这种算法:
看起来可能有多个字符串组成“B”(从第3行开始),因此您可以按照上面的步骤2不断迭代字符串,将字符串添加到列表或字符串生成器中(视情况而定)。我的注释嵌入到
processInput
方法中
public static void main(String[] args)
{
String input = "1: In: ThisName (ThisName) is in this list\n" +
"2: In: OtherName (With These) (OtherName (With These)) is in this list\n" +
"3: In: Really Annoying (Because) Separators (Really Annoying (Because) Separators) is in this list\n" +
"4: In: Not the Same (NotTheSame) is in this list\n" +
"5: In: A = (B) (A = (B)) is in this list\n" +
"6: In: A != (B) (A != B) is in this list\n";
for (String line : input.split("\n"))
{
processInput(line);
}
}
public static void processInput(String line)
{
// Parse the relevant part from the input.
Matcher inputPattern = Pattern.compile("(\\d+): In: (.*) is in this list").matcher(line);
if (!inputPattern.matches())
{
System.out.println(line + " is not valid input");
return;
}
String inputNum = inputPattern.group(1);
String aAndB = inputPattern.group(2);
// Check if a = b.
Matcher aEqualsBPattern = Pattern.compile("(.*) \\(\\1\\)").matcher(aAndB);
if (aEqualsBPattern.matches())
{
System.out.println("Input " + inputNum + ":");
System.out.println("a = b = " + aEqualsBPattern.group(1));
System.out.println();
return;
}
// Check if a and b have no parentheses.
Matcher noParenthesesPattern = Pattern.compile("([^()]*) \\(([^()]*)\\)").matcher(aAndB);
if (noParenthesesPattern.matches())
{
System.out.println("Input " + inputNum + ":");
System.out.println("a = " + noParenthesesPattern.group(1));
System.out.println("b = " + noParenthesesPattern.group(2));
System.out.println();
return;
}
// a and b have one or more parentheses in them.
// All you can do now is guess what a and b are.
// There is at least one " (" in the string.
String[] split = aAndB.split(" \\(");
for (int i = 0; i < split.length - 1; i++)
{
System.out.println("Possible Input " + inputNum + ":");
System.out.println("possible a = " + mergeParts(split, 0, i));
System.out.println("possible b = " + mergeParts(split, i + 1, split.length - 1));
System.out.println();
}
}
private static String mergeParts(String[] aAndBParts, int startIndex, int endIndex)
{
StringBuilder s = new StringBuilder(getPart(aAndBParts, startIndex));
for (int j = startIndex + 1; j <= endIndex; j++)
{
s.append(" (");
s.append(getPart(aAndBParts, j));
}
return s.toString();
}
private static String getPart(String[] aAndBParts, int j)
{
if (j != aAndBParts.length - 1)
{
return aAndBParts[j];
}
return aAndBParts[j].substring(0, aAndBParts[j].length() - 1);
}
我的注释嵌入在
processInput
方法中
public static void main(String[] args)
{
String input = "1: In: ThisName (ThisName) is in this list\n" +
"2: In: OtherName (With These) (OtherName (With These)) is in this list\n" +
"3: In: Really Annoying (Because) Separators (Really Annoying (Because) Separators) is in this list\n" +
"4: In: Not the Same (NotTheSame) is in this list\n" +
"5: In: A = (B) (A = (B)) is in this list\n" +
"6: In: A != (B) (A != B) is in this list\n";
for (String line : input.split("\n"))
{
processInput(line);
}
}
public static void processInput(String line)
{
// Parse the relevant part from the input.
Matcher inputPattern = Pattern.compile("(\\d+): In: (.*) is in this list").matcher(line);
if (!inputPattern.matches())
{
System.out.println(line + " is not valid input");
return;
}
String inputNum = inputPattern.group(1);
String aAndB = inputPattern.group(2);
// Check if a = b.
Matcher aEqualsBPattern = Pattern.compile("(.*) \\(\\1\\)").matcher(aAndB);
if (aEqualsBPattern.matches())
{
System.out.println("Input " + inputNum + ":");
System.out.println("a = b = " + aEqualsBPattern.group(1));
System.out.println();
return;
}
// Check if a and b have no parentheses.
Matcher noParenthesesPattern = Pattern.compile("([^()]*) \\(([^()]*)\\)").matcher(aAndB);
if (noParenthesesPattern.matches())
{
System.out.println("Input " + inputNum + ":");
System.out.println("a = " + noParenthesesPattern.group(1));
System.out.println("b = " + noParenthesesPattern.group(2));
System.out.println();
return;
}
// a and b have one or more parentheses in them.
// All you can do now is guess what a and b are.
// There is at least one " (" in the string.
String[] split = aAndB.split(" \\(");
for (int i = 0; i < split.length - 1; i++)
{
System.out.println("Possible Input " + inputNum + ":");
System.out.println("possible a = " + mergeParts(split, 0, i));
System.out.println("possible b = " + mergeParts(split, i + 1, split.length - 1));
System.out.println();
}
}
private static String mergeParts(String[] aAndBParts, int startIndex, int endIndex)
{
StringBuilder s = new StringBuilder(getPart(aAndBParts, startIndex));
for (int j = startIndex + 1; j <= endIndex; j++)
{
s.append(" (");
s.append(getPart(aAndBParts, j));
}
return s.toString();
}
private static String getPart(String[] aAndBParts, int j)
{
if (j != aAndBParts.length - 1)
{
return aAndBParts[j];
}
return aAndBParts[j].substring(0, aAndBParts[j].length() - 1);
}
您可以解析文本,但不能使用正则表达式,并且至少要满足以下条件之一:
)((
,:-)
)等Hello(-::)
,您也知道第二个Hello
之前的(
是“正确”的括号isMatchedBranchers(String)
方法,检查所有括号是否正确匹配。从零开始设置一个计数器,并扫描整个字符串
- 对于字符串中的每个字符:
- 如果当前字符是
,(
)计数器+
- 如果当前字符为
,)
计数器--
- 如果计数器为负数,则返回false
- 如果当前字符是
- 如果最后计数器为正,则返回false。否则返回true
- 查找最右边的索引(使用
)lastIndexOf
计数器=0
- 对于从该索引向下到4的每个字符(中
:之后的字符:
- 如果是
,)
计数器+++
- 如果是
,(
计数器--
- 如果
停止,则返回当前索引计数器==0
- 如果是
(
)前面的空格。你的B是从这个索引+1到你首先找到的右边)
的索引
回退方法
假设你的括号不平衡,你能做些什么吗
- 列出字符串中
的所有索引(
- 如果列表的长度为偶数-坏字符串,请向用户报告
- 如果长度为奇数,则取中间
)的索引。假设A和B相同,它们应具有相同的(
)编号,因此左侧和右侧具有相同编号的(
)是您的候选项(
- 像以前一样提取A和B。如果它们不相等-错误字符串,则向用户报告
)((
,:-)
)等Hello(-:)
,您也知道t之前的(
)