递归函数中Javahelper方法读取字符串

Question

我遇到了一个问题，在我已经工作的递归函数中添加了一个助手方法，只使用了2个参数，当添加第三个（助手方法）时，我的代码中断并寻找解决方案。该程序使用扫描仪对字符串进行键盘输入，对字符进行另一次输入，然后输出字母的出现次数。该错误发生在第二条if语句和两条return语句上。在第二次键盘输入后，我发现了错误：

线程"main"中的异常java.lang.StringIndexOutOfBoundsException：超出范围的字符串索引：-1

import java.util.Scanner;

public class recursiveString {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        System.out.println("Enter a string: ");
        String input = sc.nextLine();
        System.out.println("Enter a character to find number of occurences: ");
        char character = sc.next().charAt(0);
        System.out.println(character + " occurred " + count(input, character, input.length() - 1) + " times.");

    }

    public static int count(String str, char a, int high) {

        if (str.length() == high) // set equal to high to stop the recursion from infinitely looping
            return high;
        if (str.charAt(str.length() - 1) != a) // if the character in the string is  not equal to "a" subtract from count(substring)
            return count(str.substring(0, str.length() - 1), a, high - 1);
        else 
            return 1 + count(str.substring(0, str.length() - 1), a, high - 1); 
            // else add +1 to count for each instance of "a" in the string

    }

}

Answer 1

这里有一个可能的解决方案，可以帮助您避免索引越界：

public static int count(String str, char a, int high) {

    if (str == null || str.length() == 0) {
    // just to be extra safe, if we have an empty string or null 
        return 0;

    }
    //changed this end condition - now high describes how many steps we take before returning the answer
    if (high == 0) // to stop the recursion from infinitely looping
        return high;
    if (str.charAt(str.length() - 1) != a) // if the last character in the string is not equal to "a" subtract from count(substring)
        return count(str.substring(0, str.length() - 1), a, high - 1);
    else 
        return 1 + count(str.substring(0, str.length() - 1), a, high - 1); 
        // else add +1 to count for each instance of "a" in the string

}

Answer 2

您缺少递归方法的设计：首先，您应该关注一个问题，并为基本情况或多个情况定义它。

我对这个问题的看法是，基本情况是空字符串（但即使在这之前，也要确保它不是null），或者如果高设置为0。

我对high的理解是，您可以使用它来设置要检查字符a出现的字符串的多少个字符；随着字符串变得越来越大，检查会更直接，将字符a的搜索出现的含义赋予str.substring（0， high），但我试图保持它与您的代码相似。

//we'll use high to "tell" the count method how many characters it will consider into the occurrences from the end of the given string
public static int count(String str, char a, int high) {
    //if the string isn't valid or high just tells it to stop, return 0 as there can be no occurrences of a in str
    if(str == null || str.equals("") || high == 0)
      return 0;

    // if the last character in the string is not equal to a, let's just shrink the string
    if (str.charAt(str.length() - 1) != a)
        return count(str.substring(0, str.length() - 1), a, high - 1);

    // otherwise add this 1 occurrence to the ones it will find in the rest of the string
    else 
        return 1 + count(str.substring(0, str.length() - 1), a, high - 1); 
}

main中的调用将是：

System.out.println(character+ " occurred " + count(input, character, input.length()) + " times.");