Java递归代码未返回预期结果

Question

以下是不返回我预期的时间和内容的代码。我已经包含了调试消息的输出。感谢您花时间回顾我所做的一切。

package module4;

import java.nio.charset.spi.CharsetProvider;

import java.util.Scanner;

public class LabRecursive{

    public static void main(String[] args) {

        // Prompt user and call the method

        Scanner s = new Scanner(System.in);

        System.out.print("Enter a string: ");

        String var = s.next();

        System.out.println("There is/are " + vowels(var, 0, 0) + " vowel(s) in the string " + var);

        s.close();

    }

    public static int vowels(String input, int currentPos, int amount) {

        // Create the arrays of chars to hold vowels and the input

        System.out.println("Amount at beginning: " + amount);

        char[] vow = {'a', 'e', 'i', 'o', 'u'};        

        char[] entered = input.toLowerCase().toCharArray();

        int stringLength = input.length();

        System.out.println("stringlength " + stringLength);

        if (currentPos < stringLength) {

            for (int i=0; i<5; i++) {

                if (entered[currentPos] == vow[i]) {

                    amount++;

                    System.out.println("vowel: " + entered[currentPos]);

                    System.out.println("amount: " + amount);

                    continue;

                }

            }

                currentPos++;

                System.out.println("Amount before calling the recursive function: " + amount);

                System.out.println("currentPos before calling the recursive function: " + currentPos);

                vowels(input, currentPos, amount);       

        }

        System.out.println("Amount before returning the final amount: " + amount);

        return amount;

    }   

}

输入字符串：ijo

开始金额：0

字符串长度 3

元音：我

数量：1

调用递归函数前的量：1

调用递归函数前的 currentPos: 1

开始数量：1

字符串长度 3

调用递归函数前的量：1

调用递归函数前的 currentPos：2

开始数量：1

字符串长度 3

元音：o

数量：2

调用递归函数前的数量：2

调用递归函数前的 currentPos：3

开始数量：2

字符串长度 3

返回最终金额前的金额：2

返回最终金额前的金额：2

返回最终金额前的金额：1

返回最终金额前的金额：1

字符串 ijo 中有 1 个元音

Answer 1

此更改应该解决此问题。vowels(input,&nbsp;currentPos,&nbsp;amount);到amount&nbsp;=&nbsp;vowels(input,&nbsp;currentPos,&nbsp;amount);

Answer 2

提高速度并更好地使用递归方法。所有输入参数甚至可以是只读的。public static int vowels(final String input, final int currentPos, final int amount) {&nbsp; &nbsp; // Create the arrays of chars to hold vowels and the input&nbsp; &nbsp; System.out.println("Amount at beginning: " + amount);&nbsp; &nbsp; char[] vow = {'a', 'e', 'i', 'o', 'u'};&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp; &nbsp; char[] entered = input.toLowerCase().toCharArray();&nbsp; &nbsp; int stringLength = input.length();&nbsp; &nbsp; System.out.println("stringlength " + stringLength);&nbsp; &nbsp; if (currentPos < stringLength) {&nbsp; &nbsp; &nbsp; &nbsp; for (int i=0; i<5; i++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (entered[currentPos] == vow[i]) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.println("vowel: " + entered[currentPos]);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; System.out.println("amount: " + amount);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return vowels(input, currentPos + 1, amount + 1);&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; System.out.println("Amount before calling the recursive function: " + amount);&nbsp; &nbsp; &nbsp; &nbsp; System.out.println("currentPos before calling the recursive function: " + currentPos);&nbsp; &nbsp; &nbsp; &nbsp; return vowels(input, currentPos + 1, amount);&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; }&nbsp; &nbsp; System.out.println("Amount before returning the final amount: " + amount);&nbsp; &nbsp; return amount;}&nbsp; &nbsp;