For some reason when the string length is zero, it does not come out of the while loop. Can some one help me on this?
static String str1 = "";
public static void reverse(String str) {
while (str.length() > 0) {
str1 = str1 + str.charAt(str.length() - 1);
StringBuffer str_buf = new StringBuffer(str);
str = str_buf.deleteCharAt(str.length() - 1).toString();
reverse(str);
}
System.out.println("String is " + str1);
}
Replace while with if
if(str.length()>0)
Update:
The reason while fails is, after str.length() becomes 0, it hits the bottom of recursion, and control returns to the "higher" level, where str.length() is still 1. So it again calls itself.
So with while ,after it reaches 0, it will keep looping continuously between 1 and 0.
The way that you have it coded you are looping through the entire string and recursively calling the function. So for a short string "abcd" the first time through will call reverse with "abc", "ab", and "a". The reverse call to "abc" will call reverse with "ab" and "a". If you are recursively calling the function then you don't need the while loop as #sanjeev-mk suggested instead you just need the if as the exit condition.
I believe there is a more suitable code to recursively print string reverse:
public void reverseStringPrint(String inputString, int index){
if (index < (inputString.lenght() - 1))
reverseStringPrint(inputString, index + 1);
System.out.print(inputString.charAt(index);
}
Run with index = 0.
In order to get the reversed string at the output (not only print it) you may do it like this:
public String reverseString(String inputString, int index){
String restOfTheString = "";
if (index < (inputString.lenght() - 1))
restOfTheString = reverseStringPrint(inputString, index + 1);
return charAt(index) + restOfTheString;
}
Related
So, the task is to create a string that makes a progression throughout the letters of a string, returning a substring progressively longer.
For example if the input is Book, the answer would be: BBoBooBook . For the input Soup the method would return SSoSouSoup. I want to write it recursively. In my current method I receive no error but at the same time no anwer from the compiler.
public static String stringProgression(String str) {
int index = 1;
String result = "";
if (str.length() == 0) {
return "" ;
} else while (index <= str.length()); {
result = result + stringExplosion(str.substring(0, index));
index++;
}
return result;
}
In your code, you are using two different method names, stringProgression and stringExplosion.
Further, you have a while loop with a semicolon, while (index <= str.length()); which forms an empty loop. Since index doesn’t change in this empty loop, it will be an infinite loop when the condition is fulfilled.
Generally, a while loop contradicts the intent to have a recursive solution.
To find a recursive solution to a problem, you have to find the self-similarity in it. I.e. when you look at the intended result for Book, BBoBooBook, you can recognize that the beginning, BBoBoo is the right result for the string Boo, and BBo is the right result for Bo. So, the original string has to be appended to the result of a recursive evaluation of the substring:
public static String stringProgression(String str) {
if(str.isEmpty()) {
return str;
}
return stringProgression(str.substring(0, str.length() - 1)) + str;
}
An alternative, shorter syntax for the same is:
public static String stringProgression(String str) {
return str.isEmpty()? str: stringProgression(str.substring(0, str.length() - 1)) + str;
}
Check this one:
private static String doStringProgression(String str, String res, int length) {
if(length > str.length()) {
return res;
}
return doStringProgression(str, res + str.substring(0, length), length + 1);
}
And you can call the method with input like in the following example:
public static String stringProgression(String str) {
return doStringProgression(str, "", 1);
}
I am trying to figure out how to search a string for a certain number then if that number exists in the string do the following:
String str = "String4";
int myInt = 0;
public static void checkString(String str) // FIX ME
{
if(str.indexOf('3') == 0)
{
myInt = 3;
}
else if(str.indexOf('4') == 0)
{
myInt = 4;
}
else if(str.indexOf('5') == 0)
{
myInt = 5;
}
}
This never returns true though, is there an alternative way to search the string. There is a lot of extra code, but this method is the only thing causing the problem. I am pulling my hair out because of it, some help would be much appreciated!
For example
if(str.indexOf('3') == 0)
search everytime 3 at position 0, it's not appropriate, because digit can be everywhere in string.
Use instead
str.indexOf('3')!=-1
and retrieve position with return value of indexOf
Check if a substring is in the given string using the contains method.
public static void checkString(String str)
{
if(str.contains("3"))
myInt = 3;
else if(str.contains("4"))
myInt = 4;
else if(str.contains("5"))
myInt = 5;
}
Let's start with an explanation of your approach before we get to an answer.
Starting with your string String str = "String4", I'm assuming that you pass it to checkString().
The indexOf() function for a string searches the string for a substring (a character in your case) and returns its position, indexed from 0. For example, if you call str.indexOf("t") it will return 1 because the character t is in position 1. Therefore, in your code you check if the numbers 3, 4, and 5 reside in the strings index 0 (the first character).
If you want to use indexOf() for this function, you can check if the number is in the string in the first place. IndexOf() returns -1 if the character your searching for isn't in the string, so you can use the following:
if (str.indexOf("3") != -1){
//do your stuff
}
And the same for 4 and 5.
Suppose I have a method called withoutX that, when given a string, removes first and last letters of the string if they are x.
My code is:
public String withoutX(String str) {
if(str.length()>0 && str.substring(0,1).equals("x")) {
str = str.substring(1);
}
if(str.length()>0 && str.substring(str.length()-1).equals("x")) {
str = str.substring(0,str.length()-1);
}
return str;
}
Why does this return the empty string ("") when str = "x"? Wouldn't the case when str = "x" fail after the first if statement because str = str.substring(1) references an index that is out of bounds (the max. index of str = "x" is 0)?
Wouldn't the case when str = "x" fail after the first if statement because str = str.substring(1) references an index that is out of bounds (the max. index of str = "x" is 0)?
No, because it's not out of bounds: the first (and second) parameters of substring can be anything up to the length() of the string, not just length() - 1. From the Javadoc:
[Throws] IndexOutOfBoundsException - if beginIndex is negative or larger than the length of this String object.
That's larger than, not larger than or equal to.
It would be easier (and more efficient) to write this as:
int start = str.startsWith("x") ? 1 : 0;
int end = str.length() - (str.endsWith("x") && start < str.length() ? 1 : 0);
return str.substring(start, end);
The reason it's easier is that you're simply checking whether the string starts/ends with x; String provides methods to do that without constructing a substring first.
The reason it's more efficient is that it doesn't create intermediate substrings.
Let's go thorough the code step by step:
//str = "x"
if(str.length()>0 && str.substring(0,1).equals("x")) { // both true
str = str.substring(1);
//now: str = ""
}
if(str.length()>0 && str.substring(str.length()-1).equals("x")) { // str.length == 0.
//Since && does not evaluate the second parameter, everything is fine and if doesn't get called
}
return ""; //as str = ""
I'm trying to take the last three chracters of any string and save it as another String variable. I'm having some tough time with my thought process.
String word = "onetwotwoone"
int length = word.length();
String new_word = id.getChars(length-3, length, buffer, index);
I don't know how to use the getChars method when it comes to buffer or index. Eclipse is making me have those in there. Any suggestions?
Why not just String substr = word.substring(word.length() - 3)?
Update
Please make sure you check that the String is at least 3 characters long before calling substring():
if (word.length() == 3) {
return word;
} else if (word.length() > 3) {
return word.substring(word.length() - 3);
} else {
// whatever is appropriate in this case
throw new IllegalArgumentException("word has fewer than 3 characters!");
}
I would consider right method from StringUtils class from Apache Commons Lang:
http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#right(java.lang.String,%20int)
It is safe. You will not get NullPointerException or StringIndexOutOfBoundsException.
Example usage:
StringUtils.right("abcdef", 3)
You can find more examples under the above link.
Here's some terse code that does the job using regex:
String last3 = str.replaceAll(".*?(.?.?.?)?$", "$1");
This code returns up to 3; if there are less than 3 it just returns the string.
This is how to do it safely without regex in one line:
String last3 = str == null || str.length() < 3 ?
str : str.substring(str.length() - 3);
By "safely", I mean without throwing an exception if the string is nulls or shorter than 3 characters (all the other answers are not "safe").
The above code is identical in effect to this code, if you prefer a more verbose, but potentially easier-to-read form:
String last3;
if (str == null || str.length() < 3) {
last3 = str;
} else {
last3 = str.substring(str.length() - 3);
}
String newString = originalString.substring(originalString.length()-3);
public String getLastThree(String myString) {
if(myString.length() > 3)
return myString.substring(myString.length()-3);
else
return myString;
}
If you want the String composed of the last three characters, you can use substring(int):
String new_word = word.substring(word.length() - 3);
If you actually want them as a character array, you should write
char[] buffer = new char[3];
int length = word.length();
word.getChars(length - 3, length, buffer, 0);
The first two arguments to getChars denote the portion of the string you want to extract. The third argument is the array into which that portion will be put. And the last argument gives the position in the buffer where the operation starts.
If the string has less than three characters, you'll get an exception in either of the above cases, so you might want to check for that.
Here is a method I use to get the last xx of a string:
public static String takeLast(String value, int count) {
if (value == null || value.trim().length() == 0 || count < 1) {
return "";
}
if (value.length() > count) {
return value.substring(value.length() - count);
} else {
return value;
}
}
Then use it like so:
String testStr = "this is a test string";
String last1 = takeLast(testStr, 1); //Output: g
String last4 = takeLast(testStr, 4); //Output: ring
This method would be helpful :
String rightPart(String text,int length)
{
if (text.length()<length) return text;
String raw = "";
for (int i = 1; i <= length; i++) {
raw += text.toCharArray()[text.length()-i];
}
return new StringBuilder(raw).reverse().toString();
}
The getChars string method does not return a value, instead it dumps its result into your buffer (or destination) array. The index parameter describes the start offset in your destination array.
Try this link for a more verbose description of the getChars method.
I agree with the others on this, I think substring would be a better way to handle what you're trying to accomplish.
You can use a substring
String word = "onetwotwoone"
int lenght = word.length(); //Note this should be function.
String numbers = word.substring(word.length() - 3);
Alternative way for "insufficient string length or null" save:
String numbers = defaultValue();
try{
numbers = word.substring(word.length() - 3);
} catch(Exception e) {
System.out.println("Insufficient String length");
}
This method will return the x amount of characters from the end.
public static String lastXChars(String v, int x) {
return v.length() <= x ? v : v.substring(v.length() - x);
}
//usage
System.out.println(lastXChars("stackoverflow", 4)); // flow
I am practicing over the summer to try and get better and I am a little stuck on the following:
http://www.javabat.com/prob/p123384
Given a string, return a new string where the first and last chars have been exchanged.
Examples:
frontBack("code") → "eodc"
frontBack("a") → "a"
frontBack("ab") → "ba"
Code:
public String frontBack(String str)
{
String aString = "";
if (str.length() == 0){
return "";
}
char beginning = str.charAt(0);
char end = str.charAt(str.length() - 1);
str.replace(beginning, end);
str.replace(end, beginning);
return str;
}
Strings can be split into an array of chars and can be made with an array of chars. For more details on String objects, go to the Java API and click on String in the lower left pane. That pane is sorted alphabetically.
Edit: Since some people are being more thorough, I think I'll give additional details. Create a char array using String's .toCharArray() method. Take the first element and store it in a char, swap the first element with the last, and place the element you stored in a char into the last element into the array, then say:
String temp = new String(charArray);
and return that. This is assuming that charArray is your array of chars.
Rather than using the String.replace method, I'd suggest using the String.substring method to get the characters excluding the first and last letter, then concatenating the beginning and end characters.
Furthermore, the String.replace method will replace all occurrences of the specified character, and returns a new String with the said replacements. Since the return is not picked up in the code above, the String.replace calls really don't do much here.
This is because String in Java is immutable, therefore, the replace method cannot make any changes to the original String, which is the str variable in this case.
Also to add, this approach won't work well with Strings that have a length of 1. Using the approach above, a call to String.substring with the source String having a length of 1 will cause a StringIndexOutOfBoundsException, so that will also have to be taken care of as a special case, if the above approach is taken.
Frankly, the approach presented in indyK1ng's answer, where the char[] is obtained from the String and performing a simple swap of the beginning and end characters, then making a String from the modified char[] is starting to sound much more pleasant.
String instances in Java are immutable. This means that you cannot change the characters in a String; a different sequence of characters requires a new object. So, when you use the replace method, throw away the original string, and use the result of the method instead.
For this method, however, you probably want to convert the String instance to an array of characters (char[]), which are mutable. After swapping the desired characters, create a new String instance with that array.
A couple of hints:
Strings are immutable, meaning they cannot be changed. Hence, str.replace() does not change str, instead it returns a new string.
Maybe replace isn't the best... Consider frontBack("abcabc"): your function, if it were corrected, would replace 'a' with 'c' yielding "cbccbc", then 'c' with 'a' yielding "abaaba". That's not quite right!
The replace method in String actually returns a String, so if you were to insist on using replace, you'd do:
beginReplace = str.replace( beginning, end );
endReplace = beginReplace.replace( end, beginning );
return( str );
But this actually doesn't solve your specific problem, because replace replaces all occurences of a character in the string with its replacement.
For example, if my string was "apple" and I said "apple".replace( 'p', 'q' ), the resulting string would be "aqqle."
Yet another example without creating additional objects:
if (str.length() > 1) {
char[] chars = str.toCharArray();
// replace with swap()
char first = chars[0];
chars[0] = chars[chars.length - 1];
chars[chars.length - 1] = first;
str = new String(chars);
}
return str;
Edit: Performing the swap on length = 1 string is no-op.
Edit 2: dfa's change to copyValueOf did not make any sense as the Java source says in String.java: "// All public String constructors now copy the data." and the call is just delegated to a string constructor.
You could use a regex..
return str.replaceFirst("(.)(.*)(.)", "$3$2$1");
Just another, slightly different, approach, so you get a sense of the spectrum of possibilities. I commend your attention to the quick exit for short strings (instead of nesting the more-complicated processing in an if() clause), and to the use of String.format(), because it's a handy technique to have in your toolbox, not because it's notably better than regular "+" concatenation in this particular example.
public static String exchange(String s) {
int n = s.length();
if (n < 2)
return s;
return String.format("%s%s%s", s.charAt(n - 1), s.substring(1, n - 1), s.charAt(0));
}
Simple solution is:
public String frontBack(String str) {
if (str == null || str.length() == 0) {
return str;
}
char[] cs = str.toCharArray();
char first = cs[0];
cs[0] = cs[cs.length -1];
cs[cs.length -1] = first;
return new String(cs);
}
Using a character array (watch out for the nasty empty String or null String argument!)
Another solution uses StringBuilder (which is usually used to do String manupilation since String itself is immutable.
public String frontBack(String str) {
if (str == null || str.length() == 0) {
return str;
}
StringBuilder sb = new StringBuilder(str);
char first = sb.charAt(0);
sb.setCharAt(0, sb.charAt(sb.length()-1));
sb.setCharAt(sb.length()-1, first);
return sb.toString();
}
Yet another approach (more for instruction than actual use) is this one:
public String frontBack(String str) {
if (str == null || str.length() < 2) {
return str;
}
StringBuilder sb = new StringBuilder(str);
String sub = sb.substring(1, sb.length() -1);
return sb.reverse().replace(1, sb.length() -1, sub).toString();
}
Here the complete string is reversed and then the part that should not be reversed is replaced with the substring. ;)
if (s.length < 2) {
return s;
}
return s.subString(s.length - 1) + s.subString(1, s.length - 2) + s.subString(0, 1);
(untested, indexes may be of by one...
public String frontBack(String input)
{
return
input.substring(input.length() - 1) + // The last character
input.substring(1, input.length() - 1) + // plus the middle part
input.substring(0, 1); // plus the first character.
}
You can use a StringBuilder that represents "a mutable sequence of characters".
It has all methods needed to solve the problem: charAt, setCharAt, length and toString.
public String lastChars(String a, String b) {
if(a.length()>=1&&b.length()>=1){
String str = a.substring(0,1);
String str1 =b.substring(b.length()-1);
return str+str1;
}
else if(a.length()==0&&b.length()==0){
String v ="#";
String z ="#";
return v+z;
}
else if(a.length()==0&&b.length()>=1){
String s ="#";
String s1 = b.substring(b.length()-1);
return s+s1;
}
else if(a.length()>=1&&b.length()==0){
String f= a.substring(0,1);
String h = "#";
return f+h;
}
return a;
}
You can use this code:
public String frontBack(String str) {
if (str.length() <= 1)
return str;
String mid = str.substring(1, str.length()-1);
// last + mid + first
return str.charAt(str.length()-1) + mid + str.charAt(0);
}
class swap
{
public static void main(String[] args)
{
Scanner s=new Scanner(System.in);
System.out.println("no of elements in array");
int n=s.nextInt();
int a[]=new int[n];
System.out.println("Elements");
for(int i=0;i<n;i++)
{
a[i]=s.nextInt();
}
int b[]=new int[n];
for(int i=0;i<n;i++)
{
b[i]=a[i];
}
int end=n-1;
b[0]=b[end];
b[end]=a[0];
for(int i=0;i<n;i++)
{
System.out.println(b[i]);
}
}
}
if (str.length() <= 1) {
return str;
}
String mid = str.substring(1, str.length()-1);
return str.charAt(str.length()-1) + mid + str.charAt(0);
function frontBack(str: string) {
return str.slice(str.length - 1) + str.slice(1, -1) + str.slice(0, 1)
}
Slice will "cut out" the last letter. Counting the length of the string which is str.length -1, (plus) the reminder sliced string which starts at index 1 and is the last character which expressed at index -1, (plus) sliced last letter which is at index 0 through index 1.