I am trying to create a Palindrome program using recursion within Java but I am stuck, this is what I have so far:
public static void main (String[] args){
System.out.println(isPalindrome("noon"));
System.out.println(isPalindrome("Madam I'm Adam"));
System.out.println(isPalindrome("A man, a plan, a canal, Panama"));
System.out.println(isPalindrome("A Toyota"));
System.out.println(isPalindrome("Not a Palindrome"));
System.out.println(isPalindrome("asdfghfdsa"));
}
public static boolean isPalindrome(String in){
if(in.equals(" ") || in.length() == 1 ) return true;
in= in.toUpperCase();
if(Character.isLetter(in.charAt(0))
}
public static boolean isPalindromeHelper(String in){
if(in.equals("") || in.length()==1){
return true;
}
}
}
Can anyone supply a solution to my problem?
Here I am pasting code for you:
But, I would strongly suggest you to know how it works,
from your question , you are totally unreadable.
Try understanding this code. Read the comments from code
import java.util.Scanner;
public class Palindromes
{
public static boolean isPal(String s)
{
if(s.length() == 0 || s.length() == 1)
// if length =0 OR 1 then it is
return true;
if(s.charAt(0) == s.charAt(s.length()-1))
// check for first and last char of String:
// if they are same then do the same thing for a substring
// with first and last char removed. and carry on this
// until you string completes or condition fails
return isPal(s.substring(1, s.length()-1));
// if its not the case than string is not.
return false;
}
public static void main(String[]args)
{
Scanner sc = new Scanner(System.in);
System.out.println("type a word to check if its a palindrome or not");
String x = sc.nextLine();
if(isPal(x))
System.out.println(x + " is a palindrome");
else
System.out.println(x + " is not a palindrome");
}
}
Well:
It's not clear why you've got two methods with the same signature. What are they meant to accomplish?
In the first method, why are you testing for testing for a single space or any single character?
You might want to consider generalizing your termination condition to "if the length is less than two"
Consider how you want to recurse. One option:
Check that the first letter is equal to the last letter. If not, return false
Now take a substring to effectively remove the first and last letters, and recurse
Is this meant to be an exercise in recursion? That's certainly one way of doing it, but it's far from the only way.
I'm not going to spell it out any more clearly than that for the moment, because I suspect this is homework - indeed some may consider the help above as too much (I'm certainly slightly hesitant myself). If you have any problems with the above hints, update your question to show how far you've got.
public static boolean isPalindrome(String in){
if(in.equals(" ") || in.length() < 2 ) return true;
if(in.charAt(0).equalsIgnoreCase(in.charAt(in.length-1))
return isPalindrome(in.substring(1,in.length-2));
else
return false;
}
Maybe you need something like this. Not tested, I'm not sure about string indexes, but it's a start point.
I think, recursion isn't the best way to solve this problem, but one recursive way I see here is shown below:
String str = prepareString(originalString); //make upper case, remove some characters
isPalindrome(str);
public boolean isPalindrome(String str) {
return str.length() == 1 || isPalindrome(str, 0);
}
private boolean isPalindrome(String str, int i) {
if (i > str.length / 2) {
return true;
}
if (!str.charAt(i).equals(str.charAt(str.length() - 1 - i))) {
return false;
}
return isPalindrome(str, i+1);
}
Here is my go at it:
public class Test {
public static boolean isPalindrome(String s) {
return s.length() <= 1 ||
(s.charAt(0) == s.charAt(s.length() - 1) &&
isPalindrome(s.substring(1, s.length() - 1)));
}
public static boolean isPalindromeForgiving(String s) {
return isPalindrome(s.toLowerCase().replaceAll("[\\s\\pP]", ""));
}
public static void main(String[] args) {
// True (odd length)
System.out.println(isPalindrome("asdfghgfdsa"));
// True (even length)
System.out.println(isPalindrome("asdfggfdsa"));
// False
System.out.println(isPalindrome("not palindrome"));
// True (but very forgiving :)
System.out.println(isPalindromeForgiving("madam I'm Adam"));
}
}
public class palin
{
static boolean isPalin(String s, int i, int j)
{
boolean b=true;
if(s.charAt(i)==s.charAt(j))
{
if(i<=j)
isPalin(s,(i+1),(j-1));
}
else
{
b=false;
}
return b;
}
public static void main()
{
String s1="madam";
if(isPalin(s1, 0, s1.length()-1)==true)
System.out.println(s1+" is palindrome");
else
System.out.println(s1+" is not palindrome");
}
}
Some of the codes are string heavy. Instead of creating substring which creates new object, we can just pass on indexes in recursive calls like below:
private static boolean isPalindrome(String str, int left, int right) {
if(left >= right) {
return true;
}
else {
if(str.charAt(left) == str.charAt(right)) {
return isPalindrome(str, ++left, --right);
}
else {
return false;
}
}
}
public static void main(String []args){
String str = "abcdcbb";
System.out.println(isPalindrome(str, 0, str.length()-1));
}
Here are three simple implementations, first the oneliner:
public static boolean oneLinerPalin(String str){
return str.equals(new StringBuffer(str).reverse().toString());
}
This is ofcourse quite slow since it creates a stringbuffer and reverses it, and the whole string is always checked nomatter if it is a palindrome or not, so here is an implementation that only checks the required amount of chars and does it in place, so no extra stringBuffers:
public static boolean isPalindrome(String str){
if(str.isEmpty()) return true;
int last = str.length() - 1;
for(int i = 0; i <= last / 2;i++)
if(str.charAt(i) != str.charAt(last - i))
return false;
return true;
}
And recursively:
public static boolean recursivePalin(String str){
return check(str, 0, str.length() - 1);
}
private static boolean check (String str,int start,int stop){
return stop - start < 2 ||
str.charAt(start) == str.charAt(stop) &&
check(str, start + 1, stop - 1);
}
public static boolean isPalindrome(String str)
{
int len = str.length();
int i, j;
j = len - 1;
for (i = 0; i <= (len - 1)/2; i++)
{
if (str.charAt(i) != str.charAt(j))
return false;
j--;
}
return true;
}
Try this:
package javaapplicationtest;
public class Main {
public static void main(String[] args) {
String source = "mango";
boolean isPalindrome = true;
//looping through the string and checking char by char from reverse
for(int loop = 0; loop < source.length(); loop++){
if( source.charAt(loop) != source.charAt(source.length()-loop-1)){
isPalindrome = false;
break;
}
}
if(isPalindrome == false){
System.out.println("Not a palindrome");
}
else
System.out.println("Pailndrome");
}
}
String source = "liril";
StringBuffer sb = new StringBuffer(source);
String r = sb.reverse().toString();
if (source.equals(r)) {
System.out.println("Palindrome ...");
} else {
System.out.println("Not a palindrome...");
}
public class chkPalindrome{
public static String isPalindrome(String pal){
if(pal.length() == 1){
return pal;
}
else{
String tmp= "";
tmp = tmp + pal.charAt(pal.length()-1)+isPalindrome(pal.substring(0,pal.length()-1));
return tmp;
}
}
public static void main(String []args){
chkPalindrome hwObj = new chkPalindrome();
String palind = "MADAM";
String retVal= hwObj.isPalindrome(palind);
if(retVal.equals(palind))
System.out.println(palind+" is Palindrome");
else
System.out.println(palind+" is Not Palindrome");
}
}
Here is a recursive method that will ignore specified characters:
public static boolean isPal(String rest, String ignore) {
int rLen = rest.length();
if (rLen < 2)
return true;
char first = rest.charAt(0)
char last = rest.charAt(rLen-1);
boolean skip = ignore.indexOf(first) != -1 || ignore.indexOf(last) != -1;
return skip || first == last && isPal(rest.substring(1, rLen-1), ignore);
}
Use it like this:
isPal("Madam I'm Adam".toLowerCase(), " ,'");
isPal("A man, a plan, a canal, Panama".toLowerCase(), " ,'");
It does not make sense to include case insensitivity in the recursive method since it only needs to be done once, unless you are not allowed to use the .toLowerCase() method.
there's no code smaller than this:
public static boolean palindrome(String x){
return (x.charAt(0) == x.charAt(x.length()-1)) &&
(x.length()<4 || palindrome(x.substring(1, x.length()-1)));
}
if you want to check something:
public static boolean palindrome(String x){
if(x==null || x.length()==0){
throw new IllegalArgumentException("Not a valid string.");
}
return (x.charAt(0) == x.charAt(x.length()-1)) &&
(x.length()<4 || palindrome(x.substring(1, x.length()-1)));
}
LOL B-]
public static boolean isPalindrome(String p)
{
if(p.length() == 0 || p.length() == 1)
// if length =0 OR 1 then it is
return true;
if(p.substring(0,1).equalsIgnoreCase(p.substring(p.length()-1)))
return isPalindrome(p.substring(1, p.length()-1));
return false;
}
This solution is not case sensitive. Hence, for example, if you have the following word : "adinida", then you will get true if you do "Adninida" or "adninida" or "adinidA", which is what we want.
I like #JigarJoshi answer, but the only problem with his approach is that it will give you false for words which contains caps.
Palindrome example:
static boolean isPalindrome(String sentence) {
/*If the length of the string is 0 or 1(no more string to check),
*return true, as the base case. Then compare to see if the first
*and last letters are equal, by cutting off the first and last
*letters each time the function is recursively called.*/
int length = sentence.length();
if (length >= 1)
return true;
else {
char first = Character.toLowerCase(sentence.charAt(0));
char last = Character.toLowerCase(sentence.charAt(length-1));
if (Character.isLetter(first) && Character.isLetter(last)) {
if (first == last) {
String shorter = sentence.substring(1, length-1);
return isPalindrome(shorter);
} else {
return false;
}
} else if (!Character.isLetter(last)) {
String shorter = sentence.substring(0, length-1);
return isPalindrome(shorter);
} else {
String shorter = sentence.substring(1);
return isPalindrome(shorter);
}
}
}
Called by:
System.out.println(r.isPalindrome("Madam, I'm Adam"));
Will print true if palindrome, will print false if not.
If the length of the string is 0 or 1(no more string to check), return true, as the base case. This base case will be referred to by function call right before this. Then compare to see if the first and last letters are equal, by cutting off the first and last letters each time the function is recursively called.
Here is the code for palindrome check without creating many strings
public static boolean isPalindrome(String str){
return isPalindrome(str,0,str.length()-1);
}
public static boolean isPalindrome(String str, int start, int end){
if(start >= end)
return true;
else
return (str.charAt(start) == str.charAt(end)) && isPalindrome(str, start+1, end-1);
}
public class PlaindromeNumbers {
int func1(int n)
{
if(n==1)
return 1;
return n*func1(n-1);
}
static boolean check=false;
int func(int no)
{
String a=""+no;
String reverse = new StringBuffer(a).reverse().toString();
if(a.equals(reverse))
{
if(!a.contains("0"))
{
System.out.println("hey");
check=true;
return Integer.parseInt(a);
}
}
// else
// {
func(no++);
if(check==true)
{
return 0;
}
return 0;
}
public static void main(String[] args) {
// TODO code application logic here
Scanner in=new Scanner(System.in);
System.out.println("Enter testcase");
int testcase=in.nextInt();
while(testcase>0)
{
int a=in.nextInt();
PlaindromeNumbers obj=new PlaindromeNumbers();
System.out.println(obj.func(a));
testcase--;
}
}
}
/**
* Function to check a String is palindrome or not
* #param s input String
* #return true if Palindrome
*/
public boolean checkPalindrome(String s) {
if (s.length() == 1 || s.isEmpty())
return true;
boolean palindrome = checkPalindrome(s.substring(1, s.length() - 1));
return palindrome && s.charAt(0) == s.charAt(s.length() - 1);
}
Simple Solution
2 Scenario --(Odd or Even length String)
Base condition& Algo recursive(ch, i, j)
i==j //even len
if i< j recurve call (ch, i +1,j-1)
else return ch[i] ==ch[j]// Extra base condition for old length
public class HelloWorld {
static boolean ispalindrome(char ch[], int i, int j) {
if (i == j) return true;
if (i < j) {
if (ch[i] != ch[j])
return false;
else
return ispalindrome(ch, i + 1, j - 1);
}
if (ch[i] != ch[j])
return false;
else
return true;
}
public static void main(String[] args) {
System.out.println(ispalindrome("jatin".toCharArray(), 0, 4));
System.out.println(ispalindrome("nitin".toCharArray(), 0, 4));
System.out.println(ispalindrome("jatinn".toCharArray(), 0, 5));
System.out.println(ispalindrome("nittin".toCharArray(), 0, 5));
}
}
for you to achieve that, you not only need to know how recursion works but you also need to understand the String method.
here is a sample code that I used to achieve it: -
class PalindromeRecursive {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
System.out.println("Enter a string");
String input=sc.next();
System.out.println("is "+ input + "a palindrome : " + isPalindrome(input));
}
public static boolean isPalindrome(String s)
{
int low=0;
int high=s.length()-1;
while(low<high)
{
if(s.charAt(low)!=s.charAt(high))
return false;
isPalindrome(s.substring(low++,high--));
}
return true;
}
}
Related
I'm working on a short project to search a string for a specified substring using recursion.
I have tried using various strings and substrings, as well as making my code as simple as possible, but it always returns false if the substring is more than one character. (I have an accessor and mutator, as well as int i set to 0 before this method)
public boolean find(String target) {
if (i == target.length()) {
return true;
}
System.out.println(sentence);
if (sentence.length() < target.length()) {
return false;
}
if (getSentence().toLowerCase().charAt(0) == target.toLowerCase().charAt(0)) {
i++;
} else {
i = 0;
}
sentence = sentence.substring(1);
return find(target);
}
Tester code and output:
public static void main(String[] args) {
Sentence test = new Sentence("Lizard");
System.out.println(test.find("z"));
Sentence test2 = new Sentence("Seventeen");
System.out.println(test2.find("teen"));
}
Lizard
izard
zard
true
Seventeen
eventeen
venteen
enteen
nteen
teen
een
false
Your method only tests target at the first character, but you modify the sentence - e.g. you also need to modify your target when you recurse. Something like,
public boolean find(String target) {
if (i == target.length()) {
return true;
}
System.out.println(sentence);
if (sentence.length() < target.length()) {
return false;
}
if (sentence.toLowerCase().charAt(0) == target.toLowerCase().charAt(0)) {
i++;
} else {
i = 0;
}
sentence = sentence.substring(1);
return find(target.substring(1));
}
So for I am trying to make a boolean method that checks if an inputed string is a positive integer. It will return true if it is a positive integer and false if it is anything else. Here is my code:
public static boolean isPositiveInteger(String input) {
int stringLength = input.length();
int index = stringLength-1;
while(index>0) {
index--;
if(input.charAt(0) != '-' && input.charAt(index) >= '0' &&
input.charAt(index) <= '9') {
return true;
}
return false;
}
return false;
}
When the input is the string "fish33" the method will return true instead of false. Why is that?
Your while loop executes only once - return will stop execution. Moreover, you start with second to last, not with last character. Replace your code with this one:
public static boolean isPositiveInteger(String input) {
int stringLength = input.length();
int index = stringLength;
// special case when input is empty string
if (index == 0) {
return false;
}
while(index > 0) {
index--;
// if some of the characters is not digit, return false
if !(input.charAt(index) >= '0' &&
input.charAt(index) <= '9') {
return false;
}
}
// if the while loop does not find any other character, return true
return true;
}
There is no point to make so many manipulations, it can be solved in few lines
public static void main(String[] args) {
System.out.println(isPositiveInteger("1"));
System.out.println(isPositiveInteger("abc"));
}
public static boolean isPositiveInteger(String input) {
try {
Integer i = Integer.parseInt(input);
return i > 0;
}
catch(NumberFormatException nfe){
return false;
}
}
I started learning Java, currently I'm playing around with recursion.
I wanted to try and make a substring method which will substring from both sides by 1 character until we get the desired string.
I managed to do the first part but I'm having problem figuring out how to substring from the back.
n and m should be the indexes between which we want to substring (inclusive).
In this example result of method should be "bstri"
Here is my code:
public static void main(String[] args) {
String s = "substringme";
System.out.println(rec(s,2,6));
}
public static String rec(String s, int n, int m) {
if(n == 0 /* && missing 2nd part of condition */){
return s;
} else {
if(n>0){
s = s.substring(1);
n--;
}
if(/* missing condition */){
s= s.substring(0, s.length()-1);
}
return rec(s,n,m);
}
}
I would appreciate any help I can get.
So fixing your recursive method is fairly easy. We just do exactly the same as you did for n:
public static String rec(String s, int n, int m) {
if (n == 0 && m == 0) {
return s;
}
else {
if(n > 0) {
s = s.substring(1);
n--;
}
if(m > 0) {
s = s.substring(0, s.length()-1);
m--;
}
return rec(s,n,m);
}
}
The issue now is that the value of m given as input is measured from the start of the String and it would be way more convenient for us if it were measured from the end of the String.
We can introduce a new method to do this for us which acts as our entry point to the recursive method:
public static String substr(String s, int n, int m) {
final int newM = s.length() - m - 1; //-1 to be inclusive of the char
return rec(s, n, newM);
}
You would then change your main method to call substr() instead:
public static void main(String[] args) {
String s = "substringme";
System.out.println(substr(s,2,6));
}
I often find myself writing these kind of "entry point" methods when I'm using recursive methods. If you were doing this properly, substr would be your public-facing method and rec would be private.
I propose that you do m-- in the first loop because the definition of m as an index changes when you shorten s from the front.
public static String rec(String s, int n, int m) {
if(n == 0 && m == s.length() - 1){
return s;
} else {
if (n > 0) {
s = s.substring(1);
n--;
m--;
}
if (m < s.length() - 1) {
s = s.substring(0, s.length() - 1);
}
return rec(s, n, m);
}
}
What you could do is look for the difference between m and the length of the string, and cut off characters from the end of the string until it is the correct length.
import java.util.*;
public class Test {
public static void main(String[] args) {
String s = "substringme";
System.out.println(rec(s,2,6));
}
public static String rec(String s, int n, int m) {
if(n == 0 && s.length()-m < 1){
return s;
} else {
if(s.length()-m > 1){
s= s.substring(0, s.length()-1);
}
else if(n>0){
s = s.substring(1);
n--;
}
return rec(s,n,m);
}
}
}
Also shouldn't substring from 2 to 6 be "bstr", not "bstri"?
System.out.println(s.substring(2, 6)); //equals bstr
I am working on this question. It seems like that I have found the right answer and returns true but then it is overwritten by false.. Newbie in Java, sorry if it is a dummy question.. How do I just return true?
Thank you in advance
Question
Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s = "leetcode",
dict = ["leet", "code"].
Return true because "leetcode" can be segmented as "leet code".
import java.util.HashSet;
import java.util.Set;
public class Hi {
public static void main(String[] args) {
String str = "leetcode";
Set<String> set = new HashSet<String>();
set.add("leet");
set.add("code");
boolean b = wordBreak(str, set);
System.out.println("b is " + b);
}
public static boolean wordBreak(String s, Set<String> wordDict) {
if(s.length() == 0 || wordDict.isEmpty()) {
return false;
}
return helper(s, wordDict, 0);
}
public static boolean helper(String s, Set<String> wordDict, int index) {
if(index == s.length()) {
System.out.println("1 is called.. ");
return true;
}
int curIndex = index;
System.out.println("curIndex is " + curIndex);
while(index < s.length()) {
//System.out.println("s.length() is " + s.length());
curIndex++;
if(curIndex > s.length()) {
System.out.println("2 is called.. ");
//return false;
return false;
}
if(wordDict.contains(s.substring(index, curIndex))) {
System.out.println(s.substring(index, curIndex) + " curIndex is " + curIndex);
helper(s, wordDict, curIndex);
}
}
System.out.println("3 is called.. ");
return false;
}
output:
curIndex is 0
leet curIndex is 4
curIndex is 4
code curIndex is 8
1 is called..
2 is called..
2 is called..
b is false
This might not answer your question but I just mentioned an approach, and by no means I'm saying that my approach is better or more optimal.
In your code, there is no return true statement. The code does the right work but at the very end, since loop doesn't break anywhere, it always returns false. I mean you need to return true somewhere based on some condition and one of such conditions I mentioned in my below example.
private static boolean test(String str, Set<String> set) {
int i = 1;
int start = 0;
List<String> tokens = new ArrayList<String>();
while (i <= str.length()) {
String substring = str.substring(start, i);
if (set.contains(substring)) {
tokens.add(substring);
start = substring.length();
}
i++;
}
String abc = "";
for (String a : tokens) {
abc = abc + a;
}
System.out.println(abc);
if (abc.equals(str)) {
return true;
} else {
return false;
}
}
Below is the screenshot from debug trace from within debugger.
So here's my code, I want the output to be like this:
Given two numbers, is the second input a multiple of the first?
For Example:
Input:
3
6
Output:
true
public boolean multiple(int m, int n){
int i = 0;
int j = 0;
boolean check = true;
if(n%m == 0){
i++;
return check;
}
else{
j++;
return false;
}
}
When I try it I get an error, I think it's because the return statement is within the if and else statements.
The code is perfectly fine .. Error must be Somewhere else
public class Test1 {
public static void main(String[] args) {
System.out.println(multiple(3, 9));
}
public static boolean multiple(int m, int n){
int i = 0;
int j = 0;
boolean check = true;
if(n%m == 0){
i++;
return check;
}
else{
j++;
return false;
}
}
}
Output
true
here is output see IDEONE
The easiest way is to return the result of your if statement.
return n % m == 0;
I'm not sure what i/j are doing. You don't use them except to increment, but they are local to the function and get GC'd after the return. What you have now is basically this:
boolean bool = some_calculation();
if (bool == true)
{
return true;
}
else
{
return false;
}