您好，如果列表中存在代码，我正在尝试将响应生成为true或false。因此，如果字符串包含“单个括号内”的值，例如：“ ABC（Q，E，1）EEE”，但是字符串包含多个括号，例如：“ B（A，1），则我能够生成响应）AA（E，Z）EE”，我无法从中生成输出。我是编码和构建逻辑的新手，如果有人可以提供帮助，那就太好了。
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter the code you want to check: ");
String input = scan.next();
List<String> codes = new ArrayList<>();
codes.add("ABC(Q,E,1)EEE");
codes.add("ABDCE(E,Z,X)E");
codes.add("B(A,1)AAEEE");
codes.add("R(1,2,3,4,5)RT(U,M,N,B,V,H)(Q,E,R,F,G,H)(R,Z)");
codes.add("B(A,1)AA(E,Z)EE");
for (Iterator<String> i = codes.iterator(); i.hasNext(); ) {
String code = i.next();
String prefix = code.substring(0, code.indexOf("("));
String suffix = code.substring(code.indexOf(")") + 1);
String middle = code.substring(code.indexOf("(") + 1, code.indexOf(")"));
String[] var = middle.split(",");
String[] result = new String[var.length];
for (int j = 0; j < var.length; j++) {
result[j] = prefix + var[j] + suffix;
if (result[j].equals(input)) {
System.out.println("True: This code is present");
}
}
}
}
输出（有效）：
Enter the code you want to check:
BAAAEEE
True: The code is present
输出（无效）：
Enter the code you want to check:
BAAAZEE
<gives no output>
让我给您一个正在做的事的示例（“ ABC（Q，E，1）EEE”）：它使该字符串的三个可能的输出分别是：“ ABCQEEE”，“ ABCEEEE”，“ ABC1EEE” ”。因此，如果我将输入指定为“ ABCQEEE”，它将在内部生成这些输出，并且如果代码在列表中的任何位置都将其输出为True。
答案 0 :(得分：0)
尝试一下。 已编辑：添加了代码注释。
import java.util.ArrayList;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter the code you want to check: ");
String input = scan.next();
scan.close();
List<String> codes = new ArrayList<>();
codes.add("ABC(Q,E,1)EEE");
codes.add("ABDCE(E,Z,X)E");
codes.add("B(A,1)AAEEE");
codes.add("R(1,2,3,4,5)RT(U,M,N,B,V,H)(Q,E,R,F,G,H)(R,Z)");
codes.add("B(A,1)AA(E,Z)EE");
for (Iterator<String> i = codes.iterator(); i.hasNext();) {
String code = i.next();
List<String> codePossiblity = generatePossibilities(code);
// check if the input is in the list of all the possibility
for (String s : codePossiblity) {
if (s.contains(input)) {
System.out.println("True: This code is present");
}
}
}
}
/* This method removes the parenthesis and generates all the possibilities.
* This method assumes that the parenthesis always comes in pair, thus
* for every opening parenthesis ["("] there is a closing parenthesis [")"]
* Example if the "code" is [A(WX)C(YZ)] then it will generate AWCY, AWCZ, AXCY and AXCZ
*
* @param code - The string which contains parenthesis.
* @return a list of all the possibilities
*/
public static List<String> generatePossibilities(String code) {
// This will hold the left part of the possibleCodes (part before "(")
List<String> possibleCodeList = new LinkedList<>();
String s = code;
boolean first = true;
// Loop while an open parenthesis ["("] can be found
while (s.contains("(")) {
// Retrieve from the string the first substring where "(" starts and ends with ")"
// In the example, in the first iteration will be "WX"
// in the second iteration this will be "YZ"
String inside = s.substring(s.indexOf("(") + 1, s.indexOf(")"));
// Retrieve the list inside the "(" and ")"
// In the example, in the first iteration the list will have "W", "X"
// in the second iteration the list will have "Y", "Z"
String[] listOfChoices = inside.split(",");
// This will hold the right part of the possibleCodes (part after ")")
List<String> listOfCombinations = new LinkedList<>();
// Loop all the possible choices
for (String choice : listOfChoices) {
// If it is the first iteration then you need to include those characters before the "("
if (first) {
// add the characters before the "(" and the remaining characters after ")"
// In the example, the first iteration of this list ("W", "X") will add "AWC(YZ)"
// the second iteration of this list ("W", "X") will add "AXC(YZ)"
listOfCombinations.add(s.substring(0, s.indexOf("(")) + choice + s.substring(s.indexOf(")") + 1));
}
// Else just start with choice
else {
// add the remaining characters after ")"
// In the example, the first iteration of this list ("Y", "Z") will add "Y"
// the second iteration of this list ("Y", "Z") will add "Z"
listOfCombinations.add(choice + s.substring(s.indexOf(")") + 1));
}
}
// Remove the subtring before the ")", in the example this will be "C(YZ)"
s = s.substring(s.indexOf(")") + 1);
// If it is the first iteration then you just need to assign the listOfCombinations directly to possibleCodeList,
// since possibleCodeList is still empty
if (first) {
possibleCodeList = listOfCombinations;
first = false;
}
// Else combine the left and right part
else {
List<String> codePossiblity2 = new LinkedList<>();
// Iterate though all the list of possible codes since we want all the elements in the list to be concatenated with the right half of the string
// The list will have "AWC(YZ)" and "AXC(YZ)"
for (String possibleCodes : possibleCodeList) {
// Iterate the possible combinations of the right half of the original string (the second pair of "()")
// The list will have "Y" and "Z"
for (String sTmp : listOfCombinations) {
// Replace the string which are inside the "()" in the left half of the original string.
// Replace it with the right half of the original string
// In the string of "AWC(YZ)" replace "(YZ)" with "Y"
// In the string of "AWC(YZ)" replace "(YZ)" with "Z"
// In the string of "AXC(YZ)" replace "(YZ)" with "Y"
// In the string of "AXC(YZ)" replace "(YZ)" with "Z"
String t = possibleCodes.replace("(" + inside + ")", sTmp);
// add the newly created string to codePossiblity2
codePossiblity2.add(t);
}
// At the end of the loop above codePossiblity2 will have these values
// AWCY, AWCZ, AXCY and AXCZ
}
// overwrite the possibleCodeList since we have now a new left part of the string
possibleCodeList = codePossiblity2;
}
}
return possibleCodeList;
}
}
答案 1 :(得分：0)
如果您要做的只是根据用户输入正确或错误，则可以将代码字符串转换为正则表达式，并检查输入是否与正则表达式列表匹配。
步骤：
将代码列表中的每个元素转换为正则表达式
//将“ ABC（Q，E，1）EEE”转换为“ ABC [QE1] EEE”，以匹配每个以ABC开头，后跟[QE1]中的一个并以EEE结尾的字符串
//“ R（1,2,3,4,5）RT（U，M，N，B，V，H）（Q，E，R，F，G，H）（R，Z） ”转换为“ R [12345] RT [UMNBVH] [QERFGH] [RZ]”
等
检查输入是否与正则表达式之一匹配
示例：
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter the code you want to check: ");
String input = scan.next();
List<String> codes = new ArrayList<>();
codes.add("ABC(Q,E,1)EEE");
codes.add("ABDCE(E,Z,X)E");
codes.add("B(A,1)AAEEE");
codes.add("R(1,2,3,4,5)RT(U,M,N,B,V,H)(Q,E,R,F,G,H)(R,Z)");
codes.add("B(A,1)AA(E,Z)EE");
//list to store the modified strings
List<String> modifiedCodes = new ArrayList<>();
//for each string in list find if there is a pattern like '('some chars')'
Pattern p = Pattern.compile("\\(.*\\)");
for (Iterator<String> i = codes.iterator(); i.hasNext();) {
String code = i.next();
StringBuffer sb = new StringBuffer ();
Matcher m = p.matcher(code);
while (m.find()) {
String match = m.group();
//if found a match replace '(' and ')' with '[' and ']' and remove commas
m.appendReplacement(sb, match.replace('(', '[').replace(')', ']').replace(",", ""));
}
m.appendTail(sb);
//add modified string to list
modifiedCodes.add(sb.toString());
}
boolean codeIsPresent = false;
for(String code: modifiedCodes){
//check if input matches one of the regex in the list 'modifiedCodes'
if (input.matches(code)) {
codeIsPresent = true;
System.out.println("True: This code is present");
break;
}
}
if(!codeIsPresent){
System.out.println("Code not found");
}
}
编辑
我们如何从中打印字符串的所有组合的列表 它正在获得输出？说，我只有一根绳子 “ BA（1,2,3）QW（A-Z，0-9）”，我希望将其所有可能的组合
以上评论中的问题与原始帖子略有不同，如果发布新问题可能会更好。您可以使用某种树形结构创建自己的算法来解决该问题，但它可能会很杂乱无章。我建议尽可能使用generex之类的第三方天秤座。您可以下载jar from the maven repo here。使用generex，您可以拥有所有可能的组合：
public static void main(String args[]){
//change your input to a regular expression
//"BA(1,2,3)QW(A-Z,0-9)" to "BA[1-3]QW[A-Z][0-9]"
Generex generex = new Generex("BA[1-3]QW[A-Z][0-9]");
List<String> matchedStrs = generex.getAllMatchedStrings();
matchedStrs.forEach(System.out::println);
}