Question

我如何将这个java程序分解为多个方法？它首先将一个字符串存储为输入，然后遍历它以将所有数字提取到数组列表中。然后它会打印所有这些数字及其总和和产品。

public class Assignment1 {
public static void main(String[] args){

    // Creates scanner for storing input string
    Scanner numScanner = new Scanner(System.in);
    String input;
    System.out.println("Enter a string of numbers to compute their sum and product:");
    System.out.println("(Enter '.' to terminate program)");
    input = numScanner.nextLine();


    // Terminates the program if there is no input
    if (input.length() == 0){
                System.out.println("Invalid input: Not enough characters");
                System.exit(-1);
    }


    // Terminates the program if the first character is '.'
    if (input.charAt(0) == '.'){
        System.out.println("Thank you for using numberScanner!");
        System.exit(-1);
    }


    // Defines all of the variables used in the loops
    int index = 0;
    int sum = 0;
    int product = 1; 
    Integer start = null;
    int end = 0;
    ArrayList<Integer> numbers = new ArrayList<>();


    // Loop that extracts all numbers from the string and computes their sum and product
            while (index < input.length()){
                if (input.charAt(index) >= 'A' && input.charAt(index) <= 'Z' && start == null){
                    index++;
                }else if (input.charAt(index) >= '1' && input.charAt(index) <= '9' && start == null){
                    start = index;
                    index++;
                }else if (Character.isDigit(input.charAt(index))){
                    index++;
                }else{
                        end = index;
                        numbers.add(Integer.parseInt(input.substring(start,end)));
                        sum += Integer.parseInt(input.substring(start,end));
                        product *= Integer.parseInt(input.substring(start,end));
                        index++;
                        start = null;
                }   
            }


            // For the last number, the end is greater than the length of the string
            // This prints the last number without using the end
            if (index == input.length()){
            numbers.add(Integer.parseInt(input.substring(start)));
            sum += Integer.parseInt(input.substring(start));
            product *= Integer.parseInt(input.substring(start));
            index++;
            }


        // Prints the Numbers, Sum and Product beside each other
        System.out.print("Numbers: ");
        for (Object a : numbers) {
            System.out.print(a.toString() + " ");
            }
        System.out.println("Sum: " + sum + " Product: " + product);
}

}

我只是不知道如何将单个方法拆分为多个方法

Answer 1

我认为在main方法中使用I / O是可以的。我要搬出的是构建numbers列表。您可以使用接受字符串的方法getNumbers并返回该字符串中的数字列表。例如：

private static List<Integer> getNumbers(String str) {
    List<Integer> numbers = new ArrayList<>();

    for (int i = 0; i < str.length(); i++) {
        char c = str.charAt(i);

        if (Character.isDigit(c))
            numbers.add(c - '0');
    }

    return numbers;
}

现在循环遍历getNumbers()返回的列表并计算元素的总和/产品应该是微不足道的：

List<Integer> numbers = getNumbers(input);

System.out.println(numbers);

int sum = 0;
int product = 1;

for (int i : numbers) {
    sum += i;
    product *= i;
}

System.out.printf("Sum: %d Product: %d%n", sum, product);

如何将我的程序分解为多种方法？

1 个答案: