Question

在我的学校作业中，我必须制作递归方法，并且只能使用.charAt，.indexOf，.substring，.toLowerCase，.concat以及我在代码中使用的其他一些方法

这是代码

/*
 * Lab to perform different functions on Strings
 * all methods are static
 * only two methods should be public
 * all other methods are internal (only used in the class)
 */

package stringutil;

/**
 * @author [REDACTED]
 */

public class StringUtil {

    public static boolean inOut(String input){//the argument is in main
        int len = input.length();
        boolean test;

        input = input.toLowerCase();

        //call the cleaners
        input = StringUtil.cleanse(input, len);

        //this is le final product
        String reverse = StringUtil.flip(input);
        test = input.equals(reverse);

        return test;
    }

    private static String cleanse(String raw, int count){
        if (count < 0)
            return ("");
        //this means that there was invalid punctuation
        else{
            char ch;
            ch = raw.charAt(count);

            if (ch >= 97 && ch <= 122 || ch >= 48 && ch<= 57){
                //call method again with count-1 | string is same
                return cleanse(raw, count-1);
            }
            else{ //character ain't ok yo
                if (raw.indexOf(count) == -1){
                    raw = raw.substring(raw.length()-count, count-1);
                }
                else
                    raw = raw.substring(0,count-1).concat(raw.substring(count+1));
                return cleanse(raw, count);
            }
        }
    }

    public static String flip(String input){
        String newer;
        // base case
        if (input.length() == 1){
            return input;
        }
        else{
        //take the last letter and make it the new start
            newer = input.substring(input.length()-1);
            input = input.substring(0, input.length()-1);
            return newer + flip(input);
        }
        //input = newer +
        // flip(input.substring(0, input.length()-1));
    }

/**

* @param args the command line arguments

*/

    public static void main(String[] args) {
        // TODO code application logic here
        System.out.println(StringUtil.flip("aashf"));
        System.out.println(StringUtil.inOut("what, t;haw"));
    }
}

所以我在运行

后得到这个

fhsaa
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 11
    at java.lang.String.charAt(String.java:646)
    at stringutil.StringUtil.cleanse(StringUtil.java:36)
    at stringutil.StringUtil.inOut(StringUtil.java:21)
    at stringutil.StringUtil.main(StringUtil.java:78)
Java Result: 1
BUILD SUCCESSFUL (total time: 0 seconds)

我已经摆弄了清理非字母或数字的字符的方法，但系统似乎喜欢从字符串到字符的转移。

我的翻转方法有效，但我的清洁总是遇到一个超出范围的错误。我尝试添加许多东西以确保它在范围内，但这只会增加问题。

Answer 1

Yup所以.length()为一个字符串返回字符串中的一个或多个字符，所以你总是检查越界，因为你需要在得到长度之后做一个减1，以便不超出界限因为{ {1}}返回特定地点的字母

Answer 2

有几个逻辑错误;我想这就是你想要做的事情：

$(window).resize(function(){

   // Place your code here which sets the width and height of canvas.

});

基本上，您的/* * Lab to perform different functions on Strings * all methods are static * only two methods should be public * all other methods are internal (only used in the class) */ package stringutil; /** * @author [REDACTED] */ public class StringUtil { public static boolean inOut(String input) {// the argument is in main int len = input.length(); boolean test; input = input.toLowerCase(); // call the cleaners input = StringUtil.cleanse(input, len - 1); // this is le final product String reverse = StringUtil.flip(input); test = input.equals(reverse); return test; } private static String cleanse(String raw, int count) { if (count < 0) return raw; // this means that there was invalid punctuation else { char ch; ch = raw.charAt(count); if (ch >= 97 && ch <= 122 || ch >= 48 && ch <= 57) { // call method again with count-1 | string is same return cleanse(raw, count - 1); } else { // character ain't ok yo raw = raw.substring(0, count).concat(raw.substring(count + 1)); return cleanse(raw, count - 2); } } } public static String flip(String input) { String newer; // base case if (input.length() == 1) { return input; } else { // take the last letter and make it the new start newer = input.substring(input.length() - 1); input = input.substring(0, input.length() - 1); return newer + flip(input); } // input = newer + // flip(input.substring(0, input.length()-1)); } /** * * @param args * the command line arguments * */ public static void main(String[] args) { // TODO code application logic here System.out.println(StringUtil.flip("aashf")); System.out.println(StringUtil.inOut("what, t;ahw")); } }代码段存在缺陷，您对// character ain't ok yo的调用也是如此。我还没有检查出你的其余代码，但是cleanse中调用的代码部分似乎现在正在运行。此外，作为前TA，请考虑在您的代码中添加注释。

P.S。：我还将输入字符串更改为对main的调用输出true。

Answer 3

问题是在检查

之后使用此else代码块

raw = raw.substring(0,count-1).concat(raw.substring(count+1));

您正在尝试使用子字符串连接一个超出范围的位置。 count是String的长度，并且您尝试使用count + 1位置创建子字符串。

我的代码中StringIndexOutOfBoundsException的原因是什么

3 个答案: