学校项目中控制流和字符串的JAVA问题

Question

这是我的代码需要做的：

"此代码将字符串作为输入，如果字符串为空、仅包含空格或不以数字或 +/- 开头，则返回 0。否则，代码会将字符串作为修剪后的数字返回（否前导或尾随空格），数字后面没有尾随字母。”

这就是我所拥有的：

import java.util.Scanner;

public class stringManipulator 
{
    public static void main (String[]args) 
    {
        //initialize new system.in scanner object named input
        Scanner input = new Scanner(System.in);
        
        //initialize variables
        int index = 0;
        
        //prompt user for a string as input
        System.out.println("Enter a string: ");
        
        //store user input in string variable str
        String str = input.nextLine();
        
        //edit and output user inputed string
        if(str.length()==0)
        {
            System.out.println("return 0: String is empty");
        }
        else if(str.trim().length()==0)
        {
            System.out.println("return 0: String is only whitespace");
        }
        else while(index<str.length() && Character.isDigit(str.charAt(index)))
        {
            System.out.print(str.charAt(index));
            index++;
        }

        //close input scanner object
        input.close();
    }
}

我遇到了一些问题：

1. 我似乎无法让我的 while 循环在输出中保留“+”或“-”。
2. 我似乎无法让 while 循环从我的输出中修剪前导和尾随空格。
3. 我无法让程序在执行循环之前检测第一个字符是否不是数字或“+”或“-”。

欢迎提出任何建议！谢谢！

Answer 1

public class stringManipulator {

    public static void main(String... args) {
        Scanner scan = new Scanner(System.in);
        System.out.println(readNumber(scan));
    }

    public static String readNumber(Scanner scan) {
        System.out.print("Enter a string: ");
        String str = scan.nextLine();

        if (str.isEmpty())
            return "return 0: String is empty";
        if (str.trim().isEmpty())
            return "return 0: String is only whitespace";
        if (str.charAt(0) != '+' && str.charAt(0) != '-' && !Character.isDigit(str.charAt(0)))
            return "return 0: String does not start with a number or +/-";

        StringBuilder buf = new StringBuilder(str.length());

        for (int i = 0; i < str.length(); i++)
            if (i == 0 || Character.isDigit(str.charAt(i)))
                buf.append(str.charAt(i));

        return buf.toString();
    }
}

Answer 2

最好在单独的方法中实现功能，返回特定的结果以方便测试。

private static String process(String str) {
    if (null == str || str.isEmpty()) {
        return "0: empty";
    }
    str = str.trim();
    if (str.isEmpty()) {
        return "0: whitespace";
    }
    char first = str.charAt(0);
    if (!(Character.isDigit(first) || first == '+' || first == '-')) {
        return "0: not a number";
    }

    StringBuilder sb = new StringBuilder();
    for (int i = 1; i < str.length(); i++) {
        char c = str.charAt(i);
        if (Character.isDigit(c)) {
            sb.append(c);
        } else {
            break;
        }
    }
    // additional validation of inputs like +asd, -dsf
    if (sb.length() == 0 && (first == '+' || first == '-')) {
        return "0: not a signed number";
    }

    return first + sb.toString();
}

测试：

String[] tests = {
    "", "acv", "-", "+", "12", "-01", "12dgr", "12+12", "34 23", "45-34", 
    "+346", "-14", "-fe", "+hrr", "-12ggg", "+43gd"
};

Arrays.asList(tests).forEach(s -> System.out.printf("'%s' -> %s%n", s, process(s)));

输出：

'' -> 0: empty
'acv' -> 0: not a number
'-' -> 0: not a signed number
'+' -> 0: not a signed number
'12' -> 12
'-01' -> -01
'12dgr' -> 12
'12+12' -> 12
'34 23' -> 34
'45-34' -> 45
'+346' -> +346
'-14' -> -14
'-fe' -> 0: not a signed number
'+hrr' -> 0: not a signed number
'-12ggg' -> -12
'+43gd' -> +43

此外，正则表达式可用于验证输入并删除冗余部分。 假设要返回一个有效的整数，该方法可以实现如下：

private static int processReg(String str) {
    str = str.trim();
    if (str.isEmpty() || !str.matches("[-+\\d]\\d+.*")) { // does not start with digit,+,-
        return 0;
    }
    String num = str.replaceAll("[^-+0-9]", " ") // make "not-numeric" symbols blank
                    .trim()
                    .replaceAll("([-+\\d]\\d+)(.*)", "$1");  // keep the first number
    return Integer.parseInt(num);
}

测试：

Arrays.asList(tests).forEach(s -> System.out.printf("'%s' -> %d%n", s, processReg(s)));

输出不那么冗长：

'' -> 0
'acv' -> 0
'-' -> 0
'+0' -> 
'12' -> 12
'-01' -> -1
'12dgr' -> 12
'12+12' -> 12
'34 23' -> 34
'45-34' -> 45
'+346' -> 346
'-14' -> -14
'-fe' -> 0
'+hrr' -> 0
'-12ggg' -> -12
'+43gd' -> 43

Answer 3

这最终成为了我的解决方案，感谢所有反馈和帮助！

import java.util.Scanner;

public class stringManipulator 
{
    public static void main (String[]args) 
    {
        //initialize new system.in scanner object named input
        Scanner input = new Scanner(System.in);
        
        //initialize variables
        int index = 1;
        
        //prompt user for a string as input
        System.out.println("Enter a string: ");
        
        //store trimmed user input in string variable str
        String str = input.nextLine();
        str = str.trim();
        
        //test if string is empty or contains only whitespace
        if(str.length()==0)
        {
            System.out.println("Return 0: String is empty or is only whitespace");
        }
        //tests if string starts with a digit, '+', or '-'
        else if (str.charAt(0) == '+' || str.charAt(0) == '-' || Character.isDigit(str.charAt(0)))
        {   
            //prints '+', '-', or first digit of string
            System.out.print(str.charAt(0));
            
            //loop prints characters of str up to string length as long as they are digits
            while(index<str.length() && Character.isDigit(str.charAt(index)))
            {
                System.out.print(str.charAt(index));
                index++;
            }
        }
        //output if the string starts with anything other than '+', '-', or a digit
        else 
        {
            System.out.println("Return 0: String is not a number or does not begin with '+' or '-'");
        }

        //close input scanner object
        input.close();
    }
}