Java 我的代码中出现StringIndexOutOfBoundsException的原因是什么
在我的学校作业中,我必须使用递归方法,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 这是密码Java 我的代码中出现StringIndexOutOfBoundsException的原因是什么,java,indexoutofboundsexception,Java,Indexoutofboundsexception,在我的学校作业中,我必须使用递归方法,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 这是密码 /* * Lab to perform different functions on Strings * all methods are static * only two methods should be public * all other methods are internal (only
/*
* 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"));
}
}
我曾尝试过清理非字母或数字字符的方法,但系统似乎很享受从字符串到字符的转换
我的翻转方法有效,但我的清理总是在超出范围时出错。我尝试添加许多内容以确保它在范围内,但这只会增加问题。是的,因此
.length()
对于字符串,将返回字符串中的数量或字符,因此您总是检查出界,因为在获得长度后需要执行减号1,以避免超出范围,因为.charAt>()
返回特定位置的字母有几个逻辑错误;我猜这就是您试图做的:
/*
* 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"));
}
}
/*
*实验室在字符串上执行不同的函数
*所有方法都是静态的
*只有两种方法应该是公开的
*所有其他方法都是内部的(仅在类中使用)
*/
包stringutil;
/**
*@作者[修订]
*/
公共类StringUtil{
公共静态布尔inOut(字符串输入){//参数位于main中
int len=input.length();
布尔检验;
input=input.toLowerCase();
//叫清洁工
input=StringUtil.cleanse(输入,len-1);
//这是我们的最终产品
String reverse=StringUtil.flip(输入);
测试=输入。等于(反向);
回归试验;
}
私有静态字符串清理(字符串原始、整数计数){
如果(计数<0)
返回原材料;
//这意味着有无效的标点符号
否则{
char ch;
ch=原始字符(计数);
if(ch>=97&&ch=48&&ch在if检查后,问题在于此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某个位置,这超出了界限。count是字符串的长度,您正在尝试使用count+1位置创建子字符串。您知道所有内容都从0开始索引,对吗?是的;我是否在某个地方出错了?您对cleanse的第一次调用传递了输入及其长度,然后尝试执行charAt(length)
,这永远是不合法的,因为有效的索引是0…length-1
。raw.indexOf(count)
应该得到什么?我这样做了,现在它在第62行newer=input.substring(input.length()-1);
我正在调试,请给我一些时间
raw = raw.substring(0,count-1).concat(raw.substring(count+1));