Hi I am trying to convert from lower to uppercase. I know they are other easier ways to do this, but I want something else. It seems that the program prints out the users input more than once, so I am almost 100% sure it's the loop. But I can't find where the problem is.
String a = input.nextLine();
String c = "";
int b = a.length();
for (int i = 0 ; i < b; i++)
{
if (a.charAt(i) >= 97 && a.charAt(i) <= 122)
{
c = c + a;
System.out.println(c.toUpperCase());
}
}
Actually your code logic wasn't totally right, I mean why would you add the content of a to c? Doesn't make sense, and you excluded the letters a and z in your if.
String a = input.nextLine();
String c = "";
int b = a.length();
for (int i = 0; i < b; i++) {
if (a.charAt(i) >= 97 && a.charAt(i) <= 122) {
c = c + a.charAt(i);
}
}
System.out.println(c.toUpperCase());
You're printing c in each iteration of the loop where the if is entered instead of just once, after it ends.
Why bother going through the loop at all?
String a = input.nextLine();
System.out.println( a.toUpperCase();
will do what you want
If you want to invert the characters:
StringBuffer sb = new StringBuffer(a);
for ( int i=0; i < sb.length(); i++ ) {
char c = sb.charAt(i);
if ( c.isLower() ) {
sb.setCharAt(i,c.toUpper());
} else if ( c.isUpper() ) {
sb.setCharAt(i,c.toLower());
}
}
String result = sb.toString();
This solution should work:
for (int i = 0 ; i < b; i++)
{
if (a.charAt(i) > 97 && a.charAt(i) < 122)
{
a = a.toUpperCase();
}
}
System.out.println(a);
Related
This question already has answers here:
What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it?
(26 answers)
Closed 4 years ago.
I have been trying to find my own way of making a Rot13 algorithm in java, but when I try a phrase, it gives me this error:
java.lang.ArrayIndexOutOfBoundsException: 41
So this is my code:
:Update with whole names,
public class Rot13
{
char[] translated;
String abc = "abcdefghijklmnopjrstuvwxyzabcdefghijklmnopqrstuvwxyz";
public String ROT13(String input){
input = input.toLowerCase();
char[] sentence = input.toCharArray();
char[] ABC = abc.toCharArray();
int x = input.length();
int y = 0;
char[] translated = new char[x];
for(int i = 0; i<x;i++){
int z = 0;
if(sentence[i] == ' '){
translated[i] = ' ';
}
else {
while(y==0){
if (sentence[i] == ABC[z]){
y =1;
}
else{
z += 1;
}
}
translated[i] = ABC[z+12];
}
}
String rot13string = new String(translated);
return rot13string;
}
}
: Update 2
I just tested this version again and it translates the it. But in the wrong way, for example, "Hello" becomes "tmmmm" The first letter seems to be right but then the next ones are always 'm'.
Update 3: Thanks for your answers guys, here is my final code, I just duplicated my alphabet a "few" times. :
public class ROT13
{
char[] translated;
String ab = "abcdefghijklmnopjrstuvwxyzabcdefghijklmnopqrstuvwxyz";
String abc = String.format("%0" + 1000 + "d", 0).replace("0",ab);
public String ROT13(String input){
input = input.toLowerCase();
char[] sentence = input.toCharArray();
char[] ABC = abc.toCharArray();
int length = input.length();
char[] translated = new char[length];
for(int i = 0; i<length;i++){
int y = 0;
int h = 0;
if(sentence[i] == ' '){
translated[i] = ' ';
}
else {
while(y==0){
int z = 0;
if (sentence[i] == ABC[h]){
y +=1;
}
else{
z += 1;
h += 1;
}
}
translated[i] = ABC[h+13];
}
}
String rot13string = new String(translated);
return rot13string;
}
}
Currently, think about what happens when there is a 'z' in the sentence. Then, you make t[i] = ABC[z (26) + 12] //which is larger than ABC's length.
Personally, I would do rot13 like so:
public char rot13(char s){
if (c >= 'a' && c <= 'm') return c += 13;
else if (c >= 'A' && c <= 'M') return c += 13;
else if (c >= 'n' && c <= 'z') return c -= 13;
else if (c >= 'N' && c <= 'Z') return c -= 13;
else return c;
}
How can I add 0 in front of every single digit number? I mean 1 to 01 etc.
I have tried to add ifs like
if(c >='A' && c<= 'I')
str = "0"+str;
but it just adds 0 in front of everything like abcd converts to 00001234 not 01020304.
This is my code.
String A[] = new String[size];
for (int i = 0; i < size; i++) {
A[i] = jList1.getModel().getElementAt(i);
String[] Text = A[i].split("");
String s = jList1.getModel().getElementAt(i);
String str = ("");
for (int z = 0; z < Text.length; z++) {
for (int y = 0; y < Text[z].length(); y = y + 1) {
char c = s.charAt(z);
if (c >= 'A' && c <= 'Z') {
str += c - 'A' + 1;
} else if (c >= 'a' && c <= 'z') {
str += c - 'a' + 1;
} else {
str += c;
}
}
str = str + "";
}
}
This Worked for me
public String addZero(int number)
{
return number<=9?"0"+number:String.valueOf(number);
}``
One way to do this would be to use a StringJoiner with Java 8:
String s = "abcdABCD";
s = s.chars()
.mapToObj(i -> Integer.toString((i >= 'a' && i <= 'z' ? i - 'a' : i - 'A') + 1))
.collect(Collectors.joining("0", "0", "")));
System.out.println(s);
>> 0102030401020304
String str = "abcd-zzz-AAA";
StringBuilder sb = new StringBuilder();
for (int i = 0; i < str.length(); i++) {
char ch = str.toLowerCase().charAt(i);
if (ch >= 'a' && ch <= 'z') {
sb.append('0');
sb.append(ch - 'a' + 1);
} else {
sb.append(ch);
}
}
Result: abcd-zzz-AAA -> 01020304-026026026-010101
Final fix :-)
use String#chars to get a stream of its characters, then for each one do the manipulation you want.
public class Example {
public static void main(String[] args) {
String s = "aBcd1xYz";
s.chars().forEach(c -> {
if (c >= 'a' && c <= 'z')
System.out.print("0" + (c - 'a' + 1));
else if (c >= 'A' && c <= 'Z')
System.out.print("0" + (c - 'A' + 1));
else
System.out.print(c);
});
}
}
Ouput:
0102030449024025026
You can add zero in front of single digit number using String.format.
System.out.println(String.format("%02d",1));
System.out.println(String.format("%02d",999));
The first line will print 01, second line prints 999 no zero padding on the left.
Padding zero with length of 2 and d represents integer.
I hope this helps.
I have to count the number of distinct characters alphabet in the string, so in this case the count will be - 3 (d, k and s).
Given the following String:
String input;
input = "223d323dk2388s";
count(input);
My Code :
public int count(String string) {
int count=0;
String character = string;
ArrayList<Character> distinct= new ArrayList<>();
for(int i=0;i<character.length();i++){
char temp = character.charAt(i);
int j=0;
for( j=0;j<distinct.size();j++){
if(temp!=distinct.get(j)){
break;
}
}
if(!(j==distinct.size())){
distinct.add(temp);
}
}
return distinct.size();
}
Output : 3
Are there any native libraries which return me the number of characters present in that string ?
With java 8 it is much easy. You could use something like this
return string.chars().distinct().count();
One way is to maintain an array and then fill it up and get the total. This checks for all characters including special characters and numbers.
boolean []chars = new boolean[256];
String s = "223d323dk2388s";
for (int i = 0; i < s.length(); ++i) {
chars[s.charAt(i)] = true;
}
int count = 0;
for (int i = 0; i < chars.length; ++i) {
if (chars[i]) count++;
}
System.out.println(count);
Here's an alternative if you want to calculate the count only of letters, not including numbers and special symbols. Note that capital and small alphabets are different.
boolean []chars = new boolean[56];
String s = "223d323dk2388szZ";
for (int i = 0; i < s.length(); ++i) {
char ch = s.charAt(i);
if (ch >=65 && ch <= 90) {
chars[ch - 'A'] = true;
} else if (ch >= 97 && ch <= 122) {
chars[ch - 'a' + 26] = true; //If you don't want to differentiate capital and small differently, don't add 26
}
}
int count = 0;
for (int i = 0; i < chars.length; ++i) {
if (chars[i]) count++;
}
System.out.println(count);
Another way of doing it is using a Set.
String s = "223d323dk2388s";
Set<Character> set = new HashSet<Character>();
for (int i = 0; i < s.length(); ++i) {
set.add(s.charAt(i));
}
System.out.println(set.size());
If you don't want numbers and special symbols.
String s = "223d323dk2388s";
Set<Character> set = new HashSet<Character>();
for (int i = 0; i < s.length(); ++i){
char ch = s.charAt(i);
if ((ch >= 65 && ch <= 90) || (ch >= 97 && ch <= 122))
set.add(s.charAt(i));
}
System.out.println(set.size());
String s="InputString";
String p="";
char ch[]=s.toCharArray();
for(int i=0;i<s.length();i++)
{
for(int j=i+1;j<s.length();j++)
{
if(ch[i]==ch[j])
{
ch[j]=' ';
}
}
p=p+ch[i];
p = p.replaceAll("\\s","");
}
System.out.println(p.length());
To those coming here looking for a solution in Kotlin:
val countOfDistinctCharacters = string.toCharArray().distinct().size
(EDITED)
My problem statement: write a method that will encode the String passed to the method by adding 13 letters to each character in the String. If the letter after adding 13 exceeds 'z' then "wrap around" the alphabet. Then return the encoded String.
encodeString("hello") → "uryyb"
encodeString("pie") → "cvr"
encodeString("book") → "obbx"
this is what I have so far :
public static String encodeString (String input) {
String output;
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (c >= 'a' && c <= 'm')
c += 13;
else if (c >= 'n' && c <= 'z')
c -= 13;
output= (" " + (c));
}
return output;
}
now I know that I have to create a counter so that the method will continue to loop until it reaches the length of the string passed...and I know that if the charAt(index) is less than the character 'n' that I add 13 and if it is greater then I subtract 13. when I put it all together though I just get so confused and just get a bunch of compiling errors like Type mismatch: cannot convert from int to String.
note straightforward explanations/answers would be much appreciated...
***so now my problem is that it keeps telling me my output variable may not have been initialized
This code is not the most performatic but works good with Upper and Lower characters.
hElLo → uRyYb
pIe → cVr
bOoK → oBbX
private static String encodeString(String string) {
char[] ret = new char[string.length()];
for (int i = 0; i < string.length(); i++) {
ret[i] = rot13(string.charAt(i));
}
return String.valueOf(ret);
}
public static char rot13(char c) {
if (Character.isLetter(c)) {
if (Character.compare(Character.toLowerCase(c), 'a') >= 0
&& Character.compare(Character.toLowerCase(c), 'm') <= 0)
return c += 13;
else
return c -= 13;
}
return c;
}
You have to initialize your output variable as an empty String. Furthermore you are always replacing the contents of the output variable with the last char you've just encoded. So you have to add every char to the output with += instead of =.
So here is the fixed solution:
public static String encodeString(String input) {
String output = ""; // initialize as empty String
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (c >= 'a' && c <= 'm') {
c += 13;
} else if (c >= 'n' && c <= 'z') {
c -= 13;
}
output += " " + c; // add all chars to the String instead of replacing the whole String with "="!
}
return output;
}
I beautified your code a bit, so everybody can see what it really does.
Use an IDE!
I want to evaluate an expression like -4-12-2*12-3-4*5 given in a String form without using API as I am a beginner and want to grasp the logic.
Given below is my unsuccessful attempt to this problem which, if you may like, ignore and suggest appropriate logic.And of course your codes are also welcome :-)
public class SolveExpression3 {
static String testcase1 = "-4-12-2*12-3-4*5";
public static void main(String args[]){
SolveExpression3 testInstance= new SolveExpression3();
int result = testInstance.solve(testcase1);
System.out.println("Result is : "+result);
}
public int solve(String str){
int sum = 1;
int num1 = 0;
int num2 = 0;
String num = "";
int len = str.length();
System.out.println(str);
for (int i = len-1 ; i >= 0; i--)
{
char ch = str.charAt(i);
if(ch == '*')
{
String s = "";
num1 = num2 = 0;
//to get the number on left of *
for (int j = i; j >= 0; j--)
{
char c = str.charAt(j);
if(c == '+' || c == '-' || j == 1)
{
num1 = stringToInt(s);
s = "";
break;
}
else
{
s = c + s;
}
}
//to get the number on right of *
for (int j = i; j <= len; j++)
{
char c = str.charAt(j);
if(c == '+' || c == '-' || j == len-1)
{
num2 = stringToInt(s);
s = "";
break;
}
else
{
s = c + s;
}
}
sum = sum + num1*num2;
}
else
{
num = ch + num;
}
}
len = str.length();
for (int i = len-1; i >= 0; i--)
{
char ch = str.charAt(i);
if(ch==' ')
{}
else if(ch=='+')
{
sum = sum + stringToInt(num);
num = "";
}
else if(ch=='-')
{
sum = sum - stringToInt(num);
num = "";
}
else
{
num = ch + num;
}
}
return sum;
}
public int stringToInt(String str)
{
int number=0;
for(int i = 0; i < str.length(); i++)
{
int num = str.charAt(i) - 48;
number = number*10+num;
}
return number;
}
}
found=true;
static String testcase1 = "-4-12-2*12-3-4*5";
Pattern SEGMENT_PATTERN = Pattern.compile("(\\d+(\\.\\d+)?|\\D+)");
/*\\d-means digit,
\\.-point,
+-one or more times,
?-optional and
\\D-non digit ch*/
Matcher matcher = SEGMENT_PATTERN.matcher(testcase1);
while (found) {
boolean Found = matcher.find();
String segment = matcher.group();//representing a number or an operator
if (Character.isDigit(segment.toCharArray()[0])) {
//is digit
}
else {
//is operator
}
}
This a a solution using a patter to determine if you have a number or and operator,u just have to adapt it a little to your case to computing the result.
You can add all the matches found to an array list than traverse it and test the operators and computer the result.
It works for floating numbers too,ex:"it matches 5.10".
I would suggest a different logic for your purpose.
Usually the logic behind programs algorithms is not different from the logic that you will apply if you have to do the task by hand.
For an expression like your example you would usually do:
Find all the *
For each * compute the result of the operation
Repeat steps 1 and 2 for + and -
Try to implement a recursive descent parser, a tutorial depicting how a calculator can be implemented (in Python but the same concepts apply to java) can be found here http://blog.erezsh.com/how-to-write-a-calculator-in-70-python-lines-by-writing-a-recursive-descent-parser/