此方法应该用最后一个字母
切换字符串的第一个字母public static String frontBack(String str)
{
return (str.substring(str.length()-1, str.length()) + str.substring(1, str.length()-1) + str.substring(0, 1));
}
我运行代码，它编译得很好并返回正确的输出，但是我收到一条错误消息“线程中的异常”主“java.lang.StringIndexOutOfBoundsException：字符串索引超出范围：-1”。
通过一些测试，我发现造成这个问题的代码部分是str.substring（1，str.length（） - 1），但我不知道为什么。 （1，str.length（） - 1）应在范围内。
如果重要的话，传递的参数是“代码”。
供参考，整个代码是
public class Methodical {
/*
* Given two int values, returns their sum.
* Unless the two values are the same, then return double their sum.
* sumDouble(1, 2) → 3
* sumDouble(3, 2) → 5
* sumDouble(2, 2) → 8
*/
public static int sumDouble(int a, int b)
{
if(a == b)
{
return 2 * (a + b);
}
else
{
return a + b;
}
}
/*
* Given 2 ints, a and b, return true if one of them is 10
* or if their sum is 10.
* makes10(9, 10) → true
* makes10(9, 9) → false
* makes10(1, 9) → true
*/
public static boolean makes10(int a, int b)
{
if(a == 10 || b == 10 || (a+b) == 10)
{
return true;
}
else
{
return false;
}
}
/*
* We have two monkeys, a and b, and the parameters
* aSmile and bSmile indicate if each is smiling.
* We are in trouble if they are both smiling
* or if neither of them is smiling.
* Return true if we are in trouble.
* monkeyTrouble(true, true) → true
* monkeyTrouble(false, false) → true
* monkeyTrouble(true, false) → false
*/
public static boolean monkeyTrouble(boolean aSmile, boolean bSmile)
{
if((aSmile == true && bSmile == true) || (aSmile == false & bSmile == false))
{
return true;
}
else
{
return false;
}
}
/*Return true if the given non-negative number is a multiple of 3
* or a multiple of 5. Use the % "modulus" operator
* or35(3) → true
* or35(10) → true
* or35(8) → false
* */
public static boolean or35(int a)
{
if(a % 3 == 0 || a % 5 == 0)
{
return true;
}
else
{
return false;
}
}
/*
* Given a string, return a new string where "not " has been added to the front.
* However, if the string already begins with "not", return the string unchanged.
* notString("candy") → "not candy"
* notString("x") → "not x"
* notString("not bad") → "not bad"
*/
public static String notString(String str)
{
if(str.length() >= 3)
{
if(str.substring(0, 3).equals("not"))
{
return str;
}
else
{
return ("not " + str);
}
}
else
return ("not " + str);
}
/*
* Given a string, return a new string where the first and last chars have been exchanged
* frontBack("code") → "eodc"
* frontBack("a") → "a"
* frontBack("ab") → "ba"
*/
public static String frontBack(String str)
{
return (str.substring(str.length()-1, str.length()) + str.substring(1, str.length()-1) + str.substring(0, 1));
}
/*
* We'll say that a number is "teen" if it is in the range 13..19 inclusive.
* Given 3 int values, return true if 1 or more of them is/are teen.
* hasTeen(13, 20, 10) → true
* hasTeen(20, 19, 10) → true
* hasTeen(20, 10, 13) → true
*/
public static boolean hasTeen(int num1, int num2, int num3)
{
if(num1 >= 13 && num1 <= 19 || num2 >= 13 && num2 <= 19 || num3 >= 13 && num3 <= 19)
{
return true;
}
else
{
return false;
}
}
public static void main(String[] args)
{
int result;
boolean answer;
String value;
System.out.println("***Testing sumDouble***\n");
result = sumDouble(1, 2);
System.out.println("Should print 3: " + result);
result = sumDouble(3, 2);
System.out.println("Should print 5: " + result);
result = sumDouble(2, 2);
System.out.println("Should print 8: " + result +"\n");
System.out.println("***Testing makes10***\n");
answer = makes10(9, 10);
System.out.println("Should be true: " + answer);
answer = makes10(9, 9);
System.out.println("Should be false: " + answer);
answer = makes10(1, 9);
System.out.println("Should be true: " + answer + "\n");
System.out.println("***Testing monkeyTrouble***\n");
answer = monkeyTrouble(true, true);
System.out.println("Should be true: " + answer);
answer = monkeyTrouble(false, false);
System.out.println("Should be true: " + answer);
answer = monkeyTrouble(true, false);
System.out.println("Should be false: " + answer + "\n");
System.out.println("***Testing or35***\n");
answer = or35(3);
System.out.println("Should be true: " + answer);
answer = or35(10);
System.out.println("Should be true: " + answer);
answer = or35(8);
System.out.println("Should be false: " + answer + "\n");
System.out.println("***Testing notString***\n");
value = notString("candy");
System.out.println("Should be not candy: " + value);
value = notString("x");
System.out.println("Should be not x: " + value);
value = notString("not bad");
System.out.println("Should be not bad: " + value + "\n");
System.out.println("***Testing frontBack***\n");
value = frontBack("code");
System.out.println("Should be eodc: " + value);
value = frontBack("a");
System.out.println("Should be a: " + value);
value = frontBack("ab");
System.out.println("Should be ba: " + value + "\n");
System.out.println("***Testing hasTeen***\n");
answer = hasTeen(13, 20, 10);
System.out.println("Should be true: " + answer);
answer = hasTeen(20, 19, 10);
System.out.println("Should be true: " + answer);
answer = hasTeen(20, 10, 13) ;
System.out.println("Should be true: " + answer);
answer = hasTeen(20, 10, 45) ;
System.out.println("Should be false: " + answer + "\n");
System.out.println("***End of Tests***");
}
}
答案 0 :(得分：0)
我不确定这可能是你的问题。您是否检查过您的函数输入是否为空字符串（“”）？或者输入String可以是单个字符的字符串。
我建议你在输入字符串上添加验证：
public static String frontBack(String str)
{
if(str.length() <= 1) {
return str;
} else {
return (str.substring(str.length()-1, str.length()) + str.substring(1, str.length()-1) + str.substring(0, 1));
}
}
答案 1 :(得分：0)
IndexOutOfBoundsException如果begin Index小于零或begin Index＆gt; end Index OR end Index大于String的长度。
在你的情况下，我猜你传递的字符串begin Index＆gt; end Index（即长度为1的字符串）
答案 2 :(得分：0)
你必须考虑到只有一个字符的空链和链！
这是正确的代码！
public static String frontBack(String str)
{
if(str != null && str.length() > 1) {
return (str.substring(str.length()-1, str.length()) + str.substring(1, str.length()-1) + str.substring(0, 1));
} else {
return str;
}
}