用java解决问题。 Given a string, return the sum of the digits 0-9 that appear in the string, ignoring all other characters. Return 0 if there are no digits in the string. (Note: Character.isDigit(char) tests if a char is one of the chars '0', '1', .. '9'. Integer.parseInt(string) converts a string to an int.)

可以使用Java中的正则表达式来实现该问题。具体实现步骤如下：

1.使用正则表达式匹配字符串中的所有数字，可以使用`\\d`表示数字。

2.将匹配到的数字字符串转换为整数类型并累加。

3.如果没有匹配到数字，则返回0。

下面是代码实现：

public int sumDigits(String str) {
    int sum = 0;
    Pattern pattern = Pattern.compile("\\d"); // 正则表达式，匹配数字
    Matcher matcher = pattern.matcher(str);
    while (matcher.find()) { // 查找匹配到的数字，并累加
        sum += Integer.parseInt(matcher.group());
    }
    return sum;
}

使用示例：

String str = "abc123def456";
int sum = sumDigits(str);
System.out.println(sum); // 输出 21，即 1+2+3+4+5+6

相关问题

write a Java program called SumOfDigits which reads an integer between 0 and 1000 and adds all the digits in the integer.

Sure, here's the Java program to solve your problem:

import java.util.Scanner;

public class SumOfDigits {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.print("Enter an integer between 0 and 1000: ");
        int num = input.nextInt();

        int sum = 0;
        while (num > 0) {
            sum += num % 10;
            num /= 10;
        }

        System.out.println("The sum of the digits is " + sum);
    }
}

This program prompts the user to enter an integer between 0 and 1000 and then uses a while loop to add up all the digits in the integer. The program then prints out the sum of the digits.

use C program Write a method that computes the sum of the digits in an integer. Use the following method header: int sumDigits(int n)

Sure, here's a C program that implements the `sumDigits` method as described:

#include <stdio.h>

int sumDigits(int n);

int main() {
    int n;
    printf("Enter an integer: ");
    scanf("%d", &n);
    printf("The sum of digits in %d is %d\n", n, sumDigits(n));
    return 0;
}

int sumDigits(int n) {
    int sum = 0;
    while (n != 0) {
        int digit = n % 10;
        sum += digit;
        n /= 10;
    }
    return sum;
}

The `sumDigits` method takes an integer `n` as input, and returns the sum of the digits in `n`. The main function prompts the user to input an integer, calls `sumDigits` to compute the sum of its digits, and prints the result. The `sumDigits` method uses a while loop to iterate over each digit in `n`, adding it to a running total until all digits have been processed.