there. I faced a bit trouble with java programming
**Could someone give me a hint which of the method use in order to
string= "23578893762467290465" convert to array y= [2,3,5,7]....??**
I mean, that for example y[0]=2, y[1]=3, y[2]=5....
Thanks in advance
String.toCharArray()
From the official documentation:
It returns a newly allocated character array whose length is the
length of this string and whose contents are initialized to contain
the character sequence represented by this string.
Convert String to an integer array.
String string= "23578893762467290465";
int[] intArray = new int[string.length()];
for(int i = 0; i<string.length(); i++){
intArray[i] = Integer.parseInt(String.valueOf(string.charAt(i)));
}
public class StringToInt{
public int[] convert(String arg){
char[] characters = arg.toCharArray();
int[] integers = new int[arg.length()];
for ( int i = 0; i < characters.length; i++ )
integers[i] = Character.getNumericValue(characters[i]);
return integers;
}
private void prettyPrint(int[] result){
System.out.print("array: ");
for(int i : result){
System.out.print(i+" ");
}
}
public static void main(String[]args){
StringToInt converter = new StringToInt();
int[] result = converter.convert(args.length > 0?args[0]: "23578893762467290465" );
converter.prettyPrint(result);
}
}
ok, people beat me to it. But this'll work.
in essence:
Convert to char-array.
call Character.getNumericValue() for each character.
Another way may be:
int[] arr = new int[string.length()];
for(int i = 0; i<string.length(); i++){
arr[i] = Character.getNumericValue(string.charAt(i));
System.out.println(arr[i] + " ") ;
}
Related
how can i print the result without brackets
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] arr = new int[n];
for(int i=0; i < n; i++){
arr[i] = in.nextInt();
}
int[] reverse =new int[n];
for (int i = 0; i < reverse.length; i++) {
reverse[i]=arr[arr.length-1-i];
}
in.close();
}
}
In java8, you can conveniently do any kind of String output using join, there you will have to do some manual reversing though:
String output = String.join(", ", arr);
You can't change the implementation of Arrays.toString() so you can't print your array like that.
Although, you can assign it to an string variable and then manipulate that string as you desire.
String str = Arrays.toString(reverse);
str = str.substring(1, str.length() - 1);
Heres my code that takes a string and returns an array of the ascii values for each character in the array in order. Compile error is 'array required, but java.lang.String found'
public class Q1E {
int[] stringToCodes(String characters){
int characterLength= length(characters);
int[] array=new int[characterLength];
for(int i=0;i<characterLength;i++) {
array[i] =(int) characters[i];
}
}
}
You can't use array syntax on a String, use character.charAt(i)instead. Also, you need to return the array at the end.
Java uses Unicode/UTF-16 for strings, not ASCII.
If want to restrict your method to processing characters in the ASCII range, it should throw an exception when it encounters one outside that range.
If you want a sequence of "character codes" (aka codepoints), you have to use the String.codePointAt() at method. Because String holds a counted sequences of UTF-16 code-units and there might be one or two code-units per codepoint, you only know that String.length() is an upper bound of the number of codepoints in advance.
public class Q1E {
int[] stringToCodes(String s) {
int[] codepoints = new int[s.length()]; // there might be fewer
int count = 0;
for(int cp, i = 0; i < s.length(); i += Character.charCount(cp)) {
cp = s.codePointAt(i);
// for debugging, output in Unicode stylized format
System.out.println(String.format(
cp < 0x10000 ? "U+%04X" : "U+%05X", cp));
codepoints[count++] = cp;
}
int[] array = java.util.Arrays.copyOf(codepoints, count);
return array;
}
}
Try it with this Wikipedia link on an English word:
stringToCodes("http://en.wikipedia.org/wiki/Résumé");
Your code appears to have a few bugs, it's String#length() and I would suggest you add a null check. Finally (since characters isn't an array), I think you want to use String#charAt(int)
int[] stringToCodes(String characters) {
int characterLength = 0;
if (characters != null) {
characterLength = characters.length();
}
int[] array = new int[characterLength];
for (int i = 0; i < characterLength; i++) {
array[i] = characters.charAt(i);
}
return array;
}
Of course, you could shorten it with a ternary
int characterLength = (characters != null) ? characters.length() : 0;
int[] array = new int[characterLength];
try this:
public class Test {
public static void main(String[] args) {
int []ascii=stringToCodes("abcdef");
for(int i=0;i<ascii.length;i++){
System.out.println(ascii[i]);
}
}
public static int [] stringToCodes(String characters){
int []ascii=new int[characters.length()];
for(int i=0;i<characters.length();i++){
ascii[i]=(int)characters.charAt(i);
}
return ascii;
}
}
My main method:
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String string1;
string1 = input.next();
LargeInteger firstInt = new LargeInteger(string1);
System.out.printf("First integer: %s \n", firstInt.display());
}
LargeInteger class:
public class LargeInteger {
private int[] intArray;
//convert the strings to array
public LargeInteger(String s) {
for (int i = 0; i < s.length(); i++) {
intArray[i] = Character.digit(s.charAt(i), 10); // in base 10
}
}
//display the strings
public String display() {
String result = "";
for (int i = 0; i < intArray.length; i++) {
result += intArray[i];
}
return result.toString();
}
}
You did not instantiate your array. You need something like:
private int[] intArray = new int[SIZE];
where size is the length of your array.
You are not initialize the array intArray, that way you are getting error, here is the complete program
import java.util.Scanner;
class TestForNull {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String string1;
string1 = input.next();
LargeInteger firstInt = new LargeInteger(string1);
System.out.printf ("First integer: %s \n", firstInt.display());
}
}
and this is LargeInteger
public class LargeInteger {
private int[] intArray;
//convert the strings to array
public LargeInteger(String s) {
intArray = new int[s.length()];
for (int i = 0; i < s.length(); i++) {
intArray[i] = Character.digit(s.charAt(i), 10); // in base 10
}
}
//display the strings
public String display() {
String result="";
for (int i = 0; i < intArray.length; i++) {
result += intArray[i];
}
return result.toString();
}
}
private int[] intArray;
Member variables are null by default, so you need to initialize this.
Most likely you want it the same size as your string:
public LargeInteger(String s) {
intArray = new int[s.length()]; // Create the actual array before you try to put anything in it
for (int i = 0; i < s.length(); i++) {
intArray[i] = Character.digit(s.charAt(i), 10); // in base 10
}
}
Or you should use a container that resizes itself, like ArrayList.
Diferent approach through Integer.parseInt
Integer.parseInt("yourInt");
To achieve your goal:
String a = "12345667788" //sample
String b = "";
int [] vecInt = new int[a.length()]; // The lack of initialization was your mistake as the above stated
for(int i=0; i< a.length(); i++)
{
b = a.substring(0,1);
a= a.substring(1);
vecInt[i] = Integer.parseInt(b);
}
Please be aware of Double, long have far higher range then Integer which might be enough in your case to avoid an array!
you forgot to initialize the array intArray
I would recommend to use a java.util.List
You forgot to initialize the array. You have written it in constructor and the variables declared in method or constructor needs to be initialize at the same time.
Note : Implementing your logic in Constructor is not recommended unless and until you dont have any other choice.
I have one string:
String arr = "[1,2]";
ie "[1,2]" is like a single String.
How do I convert this arr to int array in java?
String arr = "[1,2]";
String[] items = arr.replaceAll("\\[", "").replaceAll("\\]", "").replaceAll("\\s", "").split(",");
int[] results = new int[items.length];
for (int i = 0; i < items.length; i++) {
try {
results[i] = Integer.parseInt(items[i]);
} catch (NumberFormatException nfe) {
//NOTE: write something here if you need to recover from formatting errors
};
}
Using Java 8's stream library, we can make this a one-liner (albeit a long line):
String str = "[1, 2, 3, 4, 5, 6, 7, 8, 9, 0]";
int[] arr = Arrays.stream(str.substring(1, str.length()-1).split(","))
.map(String::trim).mapToInt(Integer::parseInt).toArray();
System.out.println(Arrays.toString(arr));
substring removes the brackets, split separates the array elements, trim removes any whitespace around the number, parseInt parses each number, and we dump the result in an array. I've included trim to make this the inverse of Arrays.toString(int[]), but this will also parse strings without whitespace, as in the question. If you only needed to parse strings from Arrays.toString, you could omit trim and use split(", ") (note the space).
final String[] strings = {"1", "2"};
final int[] ints = new int[strings.length];
for (int i=0; i < strings.length; i++) {
ints[i] = Integer.parseInt(strings[i]);
}
It looks like JSON - it might be overkill, depending on the situation, but you could consider using a JSON library (e.g. http://json.org/java/) to parse it:
String arr = "[1,2]";
JSONArray jsonArray = (JSONArray) new JSONObject(new JSONTokener("{data:"+arr+"}")).get("data");
int[] outArr = new int[jsonArray.length()];
for(int i=0; i<jsonArray.length(); i++) {
outArr[i] = jsonArray.getInt(i);
}
Saul's answer can be better implemented splitting the string like this:
string = string.replaceAll("[\\p{Z}\\s]+", "");
String[] array = string.substring(1, string.length() - 1).split(",");
try this one, it might be helpful for you
String arr= "[1,2]";
int[] arr=Stream.of(str.replaceAll("[\\[\\]\\, ]", "").split("")).mapToInt(Integer::parseInt).toArray();
You can do it easily by using StringTokenizer class defined in java.util package.
void main()
{
int i=0;
int n[]=new int[2];//for integer array of numbers
String st="[1,2]";
StringTokenizer stk=new StringTokenizer(st,"[,]"); //"[,]" is the delimeter
String s[]=new String[2];//for String array of numbers
while(stk.hasMoreTokens())
{
s[i]=stk.nextToken();
n[i]=Integer.parseInt(s[i]);//Converting into Integer
i++;
}
for(i=0;i<2;i++)
System.out.println("number["+i+"]="+n[i]);
}
Output :-number[0]=1
number[1]=2
String str = "1,2,3,4,5,6,7,8,9,0";
String items[] = str.split(",");
int ent[] = new int[items.length];
for(i=0;i<items.length;i++){
try{
ent[i] = Integer.parseInt(items[i]);
System.out.println("#"+i+": "+ent[i]);//Para probar
}catch(NumberFormatException e){
//Error
}
}
If you prefer an Integer[] instead array of an int[] array:
Integer[]
String str = "[1,2]";
String plainStr = str.substring(1, str.length()-1); // clear braces []
String[] parts = plainStr.split(",");
Integer[] result = Stream.of(parts).mapToInt(Integer::parseInt).boxed().toArray(Integer[]::new);
int[]
String str = "[1,2]";
String plainStr = str.substring(1, str.length()-1); // clear braces []
String[] parts = plainStr.split(",");
int[] result = Stream.of(parts).mapToInt(Integer::parseInt).toArray()
This works for Java 8 and higher.
In tight loops or on mobile devices it's not a good idea to generate lots of garbage through short-lived String objects, especially when parsing long arrays.
The method in my answer parses data without generating garbage, but it does not deal with invalid data gracefully and cannot parse negative numbers. If your data comes from untrusted source, you should be doing some additional validation or use one of the alternatives provided in other answers.
public static void readToArray(String line, int[] resultArray) {
int index = 0;
int number = 0;
for (int i = 0, n = line.length(); i < n; i++) {
char c = line.charAt(i);
if (c == ',') {
resultArray[index] = number;
index++;
number = 0;
}
else if (Character.isDigit(c)) {
int digit = Character.getNumericValue(c);
number = number * 10 + digit;
}
}
if (index < resultArray.length) {
resultArray[index] = number;
}
}
public static int[] toArray(String line) {
int[] result = new int[countOccurrences(line, ',') + 1];
readToArray(line, result);
return result;
}
public static int countOccurrences(String haystack, char needle) {
int count = 0;
for (int i=0; i < haystack.length(); i++) {
if (haystack.charAt(i) == needle) {
count++;
}
}
return count;
}
countOccurrences implementation was shamelessly stolen from John Skeet
String arr= "[1,2]";
List<Integer> arrList= JSON.parseArray(arr,Integer.class).stream().collect(Collectors.toList());
Integer[] intArr = ArrayUtils.toObject(arrList.stream().mapToInt(Integer::intValue).toArray());
i tried myself lot but can't get a solution so i'm asking help.
i have an string String input="---4--5-67--8-9---";
now i need to convert in into an string array which will look like:
String [][]output={{4},{5},{67},{8},{9}};
i tried with split() and
java.util.Arrays.toString("---4--5-67--8-9---".split("-+")
but can't find the desired answer. so what to do?
actually i need the value 4,5,67,8,9.but i'm not sure how to find them. i will treat the values as integer for further processing
String[] numbers = "---4--5-67--8-9---".split("-+");
String[][] result = new String[numbers.length][1];
for (int i = 0; i < numbers.length; i++) {
result[i][0] = numbers[i];
}
Update: to get rid of the initial empty value, you can get a substring of the input, like:
int startIdx = 0;
char[] chars = input.toCharArray();
for (int i = 0; i < chars.length; i ++) {
if (Character.isDigit(chars[i])) {
startIdx = i;
break;
}
}
input = input.substring(startIdx);
(or you can check them for not being empty (String.isEmpty()) when processing them later.)
First, here is the answer to your question. This code will generate a two-dimensional array where each element is an array consisting of a single numeric string.
final String input = "---4--5-67--8-9---";
// desired output: {{4},{5},{67},{8},{9}}
// First step: convert all non-digits to whitespace
// so we can cut it off using trim()
// then split based on white space
final String[] arrayOfStrings =
input.replaceAll("\\D+", " ").trim().split(" ");
// Now create the two-dimensional array with the correct size
final String[][] arrayOfArrays = new String[arrayOfStrings.length][];
// Loop over single-dimension array to initialize the two-dimensional one
for(int i = 0; i < arrayOfStrings.length; i++){
final String item = arrayOfStrings[i];
arrayOfArrays[i] = new String[] { item };
}
System.out.println(Arrays.deepToString(arrayOfArrays));
// Output: [[4], [5], [67], [8], [9]]
However, I think what you really need is an array of Integers or ints, so here is a revised solution:
final String input = "---4--5-67--8-9---";
// Convert all non-digits to whitespace
// so we can cut it off using trim()
// then split based on white space
final String[] arrayOfStrings =
input.replaceAll("\\D+", " ").trim().split(" ");
// Now create an array of Integers and assign the values from the string
final Integer[] arrayOfIntegers = new Integer[arrayOfStrings.length];
for(int i = 0; i < arrayOfStrings.length; i++){
arrayOfIntegers[i] = Integer.valueOf(arrayOfStrings[i]);
}
System.out.println(Arrays.toString(arrayOfIntegers));
// Output: [4, 5, 67, 8, 9]
// Or alternatively an array of ints
final int[] arrayOfInts = new int[arrayOfStrings.length];
for(int i = 0; i < arrayOfStrings.length; i++){
arrayOfInts[i] = Integer.parseInt(arrayOfStrings[i]);
}
System.out.println(Arrays.toString(arrayOfInts));
// Output: [4, 5, 67, 8, 9]
Whether you use the Integer or the int version really depends on whether you want to just do some math (int) or need an object reference (Integer).
String[] result = "---4--5-67--8-9---".split("-+");
int i;
for (i = 0; i < result.length; i++) {
if (result[i].length() > 0) {
System.out.println(result[i]);
}
}
gives me output:
4
5
67
8
9
public class split{
public static void main(String[] argv){
String str="---4--5-67--8-9---";
String[] str_a=str.split("-+");
}
}
This seems to working for me.
Using a regex pattern seems more natural in this case:
public class split {
public static int[] main(String input) {
ArrayList<String> list = new ArrayList() ;
Pattern pattern = Pattern.compile("[0-9]") ;
Matcher matcher = pattern.matcher(input) ;
String match = null ;
while( ( match = matcher.find() ) === true ) {
list.add(match) ;
}
String[] array = list.toArray( new String[ ( list.size() ) ]() ) ;
return array ;
}
}
String input="---4--5-67--8-9---";
Scanner scanner = new Scanner(input).useDelimiter("-+");
List<Integer> numbers = new ArrayList<Integer>();
while(scanner.hasNextInt()) {
numbers.add(scanner.nextInt());
}
Integer[] arrayOfNums = numbers.toArray(new Integer[]{});
System.out.println(Arrays.toString(arrayOfNums));
I thought the following is quite simple, although it uses List and Integer arrays, Its not that an overhead for small strings:
For simplicity, I am returning a single dimension array, but can be easily modified to return an array you want. But from your question, it seems that you just want a list of integers.
import java.util.*;
public class Test {
public static void main(String[] args) throws Throwable {
String input = "---4--5-67--8-9---";
System.out.println(split(input).length); // 5
}
public static Integer[] split(String input) {
String[] output = input.split("\\-+");
List<Integer> intList = new ArrayList<Integer>(output.length);
// iterate to remove empty elements
for(String o : output) {
if(o.length() > 0) {
intList.add(Integer.valueOf(o));
}
}
// convert to array (or could return the list itself
Integer[] ret = new Integer[intList.size()];
return intList.toArray(ret);
}
}
I might be late to the party but I figured I'd give the guava take on this.
String in = "---4--5-67--8-9---";
List<String> list = Lists.newArrayList(Splitter.on("-").omitEmptyStrings().trimResults().split(in));
System.out.println(list);
// prints [4, 5, 67, 8, 9]