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i'am new injava , in this problem i will insert a numbers of strings in a array , but the compiler give me this probleme :
PhoneNumber.java:29: error: incompatible types: String cannot be converted to boolean
while(test[i][j])
^
1 error
public class PhoneNumber{
public static void check_number(String[][] numbers, int n)
{
int i,j;
for(i = 0; i < n; i++)
{
for(j = 0; j < numbers[i].length; j++)
{
if(numbers[i][j] == "4" || numbers[i][j] == "5")
{
System.out.println("Done");
}
}
}
}
public static void main(String[] args)
{
String[][] test = new String[100][100];
Scanner number = new Scanner(System.in);
int n,i,j;
System.out.println("enter the number of numbers");
n = number.nextInt();
for(i = 0 ; i < n; i++)
{
System.out.println("enter the number " + i + 1);
j = 0;
while(test[i][j])
{
test[i][j] = number.nextLine();
j++;
}
}
check_number(test,n);
}
}
Here's the basic approach for a 1D String array with notes included:
import java.util.Scanner;
public class PhoneNumber{
public static void check_number(String[] numbers, int n)
{
for(int i = 0; i < n; i++)
{
System.out.println(numbers[i]);
//the String class method equals is best for comparison:
if(numbers[i].equals("4") || numbers[i].equals("5"))
{
System.out.println("Done");
}
}
}
public static void main(String[] args)
{
Scanner number = new Scanner(System.in);
int n;
System.out.println("enter the number of numbers");
n = number.nextInt();
//clean the scanner buffer after input especially with Int -> Line
number.nextLine();
//size your array after getting user input
String[] test = new String[n];
for(int i = 0 ; i < n; i++)
{
//parenthesis needed to get correct output for i + 1
System.out.println("enter the number " + (i + 1));
test[i] = number.nextLine();
}
check_number(test,n);
}
}
I'm writing a program that will print the unique character in a string (entered through a scanner). I've created a method that tries to accomplish this but I keep getting characters that are not repeats, instead of a character (or characters) that is unique to the string. I want the unique letters only.
Here's my code:
import java.util.Scanner;
public class Sameness{
public static void main (String[]args){
Scanner kb = new Scanner (System.in);
String word = "";
System.out.println("Enter a word: ");
word = kb.nextLine();
uniqueCharacters(word);
}
public static void uniqueCharacters(String test){
String temp = "";
for (int i = 0; i < test.length(); i++){
if (temp.indexOf(test.charAt(i)) == - 1){
temp = temp + test.charAt(i);
}
}
System.out.println(temp + " ");
}
}
And here's sample output with the above code:
Enter a word:
nreena
nrea
The expected output would be: ra
Based on your desired output, you have to replace a character that initially has been already added when it has a duplicated later, so:
public static void uniqueCharacters(String test){
String temp = "";
for (int i = 0; i < test.length(); i++){
char current = test.charAt(i);
if (temp.indexOf(current) < 0){
temp = temp + current;
} else {
temp = temp.replace(String.valueOf(current), "");
}
}
System.out.println(temp + " ");
}
How about applying the KISS principle:
public static void uniqueCharacters(String test) {
System.out.println(test.chars().distinct().mapToObj(c -> String.valueOf((char)c)).collect(Collectors.joining()));
}
The accepted answer will not pass all the test case for example
input -"aaabcdd"
desired output-"bc"
but the accepted answer will give -abc
because the character a present odd number of times.
Here I have used ConcurrentHasMap to store character and the number of occurrences of character then removed the character if the occurrences is more than one time.
import java.util.concurrent.ConcurrentHashMap;
public class RemoveConductive {
public static void main(String[] args) {
String s="aabcddkkbghff";
String[] cvrtar=s.trim().split("");
ConcurrentHashMap<String,Integer> hm=new ConcurrentHashMap<>();
for(int i=0;i<cvrtar.length;i++){
if(!hm.containsKey(cvrtar[i])){
hm.put(cvrtar[i],1);
}
else{
hm.put(cvrtar[i],hm.get(cvrtar[i])+1);
}
}
for(String ele:hm.keySet()){
if(hm.get(ele)>1){
hm.remove(ele);
}
}
for(String key:hm.keySet()){
System.out.print(key);
}
}
}
Though to approach a solution I would suggest you to try and use a better data structure and not just string. Yet, you can simply modify your logic to delete already existing duplicates using an else as follows :
public static void uniqueCharacters(String test) {
String temp = "";
for (int i = 0; i < test.length(); i++) {
char ch = test.charAt(i);
if (temp.indexOf(ch) == -1) {
temp = temp + ch;
} else {
temp.replace(String.valueOf(ch),""); // added this to your existing code
}
}
System.out.println(temp + " ");
}
This is an interview question. Find Out all the unique characters of a string.
Here is the complete solution. The code itself is self explanatory.
public class Test12 {
public static void main(String[] args) {
String a = "ProtijayiGiniGina";
allunique(a);
}
private static void allunique(String a) {
int[] count = new int[256];// taking count of characters
for (int i = 0; i < a.length(); i++) {
char ch = a.charAt(i);
count[ch]++;
}
for (int i = 0; i < a.length(); i++) {
char chh = a.charAt(i);
// character which has arrived only one time in the string will be printed out
if (count[chh] == 1) {
System.out.println("index => " + i + " and unique character => " + a.charAt(i));
}
}
}// unique
}
In Python :
def firstUniqChar(a):
count = [0] *256
for i in a: count[ord(i)] += 1
element = ""
for item in a:
if (count[ord(item)] == 1):
element = item;
break;
return element
a = "GiniGinaProtijayi";
print(firstUniqChar(a)) # output is P
public static String input = "10 5 5 10 6 6 2 3 1 3 4 5 3";
public static void uniqueValue (String numbers) {
String [] str = input.split(" ");
Set <String> unique = new HashSet <String> (Arrays.asList(str));
System.out.println(unique);
for (String value:unique) {
int count = 0;
for ( int i= 0; i<str.length; i++) {
if (value.equals(str[i])) {
count++;
}
}
System.out.println(value+"\t"+count);
}
}
public static void main(String [] args) {
uniqueValue(input);
}
Step1: To find the unique characters in a string, I have first taken the string from user.
Step2: Converted the input string to charArray using built in function in java.
Step3: Considered two HashSet (set1 for storing all characters even if it is getting repeated, set2 for storing only unique characters.
Step4 : Run for loop over the array and check that if particular character is not there in set1 then add it to both set1 and set2. if that particular character is already there in set1 then add it to set1 again but remove it from set2.( This else part is useful when particular character is getting repeated odd number of times).
Step5 : Now set2 will have only unique characters. Hence, just print that set2.
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
String str = input.next();
char arr[] = str.toCharArray();
HashSet<Character> set1=new HashSet<Character>();
HashSet<Character> set2=new HashSet<Character>();
for(char i:arr)
{
if(set1.contains(i))
{
set1.add(i);
set2.remove(i);
}
else
{
set1.add(i);
set2.add(i);
}
}
System.out.println(set2);
}
I would store all the characters of the string in an array that you will loop through to check if the current characters appears there more than once. If it doesn't, then add it to temp.
public static void uniqueCharacters(String test) {
String temp = "";
char[] array = test.toCharArray();
int count; //keep track of how many times the character exists in the string
outerloop: for (int i = 0; i < test.length(); i++) {
count = 0; //reset the count for every new letter
for(int j = 0; j < array.length; j++) {
if(test.charAt(i) == array[j])
count++;
if(count == 2){
count = 0;
continue outerloop; //move on to the next letter in the string; this will skip the next two lines below
}
}
temp += test.charAt(i);
System.out.println("Adding.");
}
System.out.println(temp);
}
I have added comments for some more detail.
import java.util.*;
import java.lang.*;
class Demo
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter String");
String s1=sc.nextLine();
try{
HashSet<Object> h=new HashSet<Object>();
for(int i=0;i<s1.length();i++)
{
h.add(s1.charAt(i));
}
Iterator<Object> itr=h.iterator();
while(itr.hasNext()){
System.out.println(itr.next());
}
}
catch(Exception e)
{
System.out.println("error");
}
}
}
If you don't want to use additional space:
String abc="developer";
System.out.println("The unique characters are-");
for(int i=0;i<abc.length();i++)
{
for(int j=i+1;j<abc.length();j++)
{
if(abc.charAt(i)==abc.charAt(j))
abc=abc.replace(String.valueOf(abc.charAt(j))," ");
}
}
System.out.println(abc);
Time complexity O(n^2) and no space.
This String algorithm is used to print unique characters in a string.It runs in O(n) runtime where n is the length of the string.It supports ASCII characters only.
static String printUniqChar(String s) {
StringBuilder buildUniq = new StringBuilder();
boolean[] uniqCheck = new boolean[128];
for (int i = 0; i < s.length(); i++) {
if (!uniqCheck[s.charAt(i)]) {
uniqCheck[s.charAt(i)] = true;
if (uniqCheck[s.charAt(i)])
buildUniq.append(s.charAt(i));
}
}
public class UniqueCharactersInString {
public static void main(String []args){
String input = "aabbcc";
String output = uniqueString(input);
System.out.println(output);
}
public static String uniqueString(String s){
HashSet<Character> uniques = new HashSet<>();
uniques.add(s.charAt(0));
String out = "";
out += s.charAt(0);
for(int i =1; i < s.length(); i++){
if(!uniques.contains(s.charAt(i))){
uniques.add(s.charAt(i));
out += s.charAt(i);
}
}
return out;
}
}
What would be the inneficiencies of this answer? How does it compare to other answers?
Based on your desired output you can replace each character already present with a blank character.
public static void uniqueCharacters(String test){
String temp = "";
for(int i = 0; i < test.length(); i++){
if (temp.indexOf(test.charAt(i)) == - 1){
temp = temp + test.charAt(i);
} else {
temp.replace(String.valueOf(temp.charAt(i)), "");
}
}
System.out.println(temp + " ");
}
public void uniq(String inputString) {
String result = "";
int inputStringLen = inputStr.length();
int[] repeatedCharacters = new int[inputStringLen];
char inputTmpChar;
char tmpChar;
for (int i = 0; i < inputStringLen; i++) {
inputTmpChar = inputStr.charAt(i);
for (int j = 0; j < inputStringLen; j++) {
tmpChar = inputStr.charAt(j);
if (inputTmpChar == tmpChar)
repeatedCharacters[i]++;
}
}
for (int k = 0; k < inputStringLen; k++) {
inputTmpChar = inputStr.charAt(k);
if (repeatedCharacters[k] == 1)
result = result + inputTmpChar + " ";
}
System.out.println ("Unique characters: " + result);
}
In first for loop I count the number of times the character repeats in the string. In the second line I am looking for characters repetitive once.
how about this :)
for (int i=0; i< input.length();i++)
if(input.indexOf(input.charAt(i)) == input.lastIndexOf(input.charAt(i)))
System.out.println(input.charAt(i) + " is unique");
package extra;
public class TempClass {
public static void main(String[] args) {
// TODO Auto-generated method stub
String abcString="hsfj'pwue2hsu38bf74sa';fwe'rwe34hrfafnosdfoasq7433qweid";
char[] myCharArray=abcString.toCharArray();
TempClass mClass=new TempClass();
mClass.countUnique(myCharArray);
mClass.countEach(myCharArray);
}
/**
* This is the program to find unique characters in array.
* #add This is nice.
* */
public void countUnique(char[] myCharArray) {
int arrayLength=myCharArray.length;
System.out.println("Array Length is: "+arrayLength);
char[] uniqueValues=new char[myCharArray.length];
int uniqueValueIndex=0;
int count=0;
for(int i=0;i<arrayLength;i++) {
for(int j=0;j<arrayLength;j++) {
if (myCharArray[i]==myCharArray[j] && i!=j) {
count=count+1;
}
}
if (count==0) {
uniqueValues[uniqueValueIndex]=myCharArray[i];
uniqueValueIndex=uniqueValueIndex+1;
count=0;
}
count=0;
}
for(char a:uniqueValues) {
System.out.println(a);
}
}
/**
* This is the program to find count each characters in array.
* #add This is nice.
* */
public void countEach(char[] myCharArray) {
}
}
Here str will be your string to find the unique characters.
function getUniqueChars(str){
let uniqueChars = '';
for(let i = 0; i< str.length; i++){
for(let j= 0; j< str.length; j++) {
if(str.indexOf(str[i]) === str.lastIndexOf(str[j])) {
uniqueChars += str[i];
}
}
}
return uniqueChars;
}
public static void main(String[] args) {
String s = "aaabcdd";
char a[] = s.toCharArray();
List duplicates = new ArrayList();
List uniqueElements = new ArrayList();
for (int i = 0; i < a.length; i++) {
uniqueElements.add(a[i]);
for (int j = i + 1; j < a.length; j++) {
if (a[i] == a[j]) {
duplicates.add(a[i]);
break;
}
}
}
uniqueElements.removeAll(duplicates);
System.out.println(uniqueElements);
System.out.println("First Unique : "+uniqueElements.get(0));
}
Output :
[b, c]
First Unique : b
import java.util.*;
public class Sameness{
public static void main (String[]args){
Scanner kb = new Scanner (System.in);
String word = "";
System.out.println("Enter a word: ");
word = kb.nextLine();
uniqueCharacters(word);
}
public static void uniqueCharacters(String test){
for(int i=0;i<test.length();i++){
if(test.lastIndexOf(test.charAt(i))!=i)
test=test.replaceAll(String.valueOf(test.charAt(i)),"");
}
System.out.println(test);
}
}
public class Program02
{
public static void main(String[] args)
{
String inputString = "abhilasha";
for (int i = 0; i < inputString.length(); i++)
{
for (int j = i + 1; j < inputString.length(); j++)
{
if(inputString.toCharArray()[i] == inputString.toCharArray()[j])
{
inputString = inputString.replace(String.valueOf(inputString.charAt(j)), "");
}
}
}
System.out.println(inputString);
}
}
I have to sort strings in an array for a school project. Our teacher won't allow us to use array,sort().
i have to use 2 sort methods but they aren't working too well.
The first one returns double of each value. ie John, jack, adam, tom will return adam,adam,jack,jack,john,john,tom,tom.
public static void sort() {
inputFileNames();//inputs list of names from a file.
for (int i = 0; i < size - 1; i++) {
for (int j = i + 1; j < size; j++) {
if (stArr[i].compareTo(stArr[j])>0) {
temp = stArr[i];
stArr[i] = stArr[j];
stArr[j] = temp;
}
}
}
display("The names are: ");// method to display array
System.out.println("");
}
the second sort doesn' run:
public static void bubbleSort() {
inputFileNames();
for (int i = size - 1; i >= 0; i--) {
for (int j = 0; j <= i; j++) {
if (stArr[j].compareTo(stArr[j+1])>0) {
temp = stArr[j];
stArr[j] = stArr[j + 1];
stArr[j + 1] = temp;
}
}
}
display("The names are: ");
System.out.println("");
}
input and display:
static void display(String heading) {
System.out.println(heading + "\n");
for (int i = 0; i < size; i++) {
System.out.println(stArr[i]);
}
}
static void inputFileNames() {
try {
Scanner scFile = new Scanner(new File("Names.txt"));
while (scFile.hasNext()) {
stArr[size] = scFile.nextLine();
size++;
}
} catch (FileNotFoundException ex) {
System.out.println("File not found.");
}
}
/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{ Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int i,j;
String[] stArr = new String[n];
for(i=0;i<n;i++)
{
stArr[i]=sc.next();
// System.out.println(stArr[i]);
}
//inputs list of names from a file.
for (i = 0; i < n ; i++) {
for (j = i+1 ; j < n; j++) {
if (stArr[i].compareTo(stArr[j])>0)
{
String temp = stArr[i];
stArr[i] = stArr[j];
stArr[j] = temp;
// System.out.println(stArr[i]);
// System.out.println(stArr[j]);
}
}
}
for(i=0;i<n;i++)
{
System.out.println(stArr[i]);
}
// your code goes here
}
}
This Is the answer for first code. I am not good in file handling so you have to use your input method. I know Scanner thats why i have used here.
In Your Second Example Your j loop is wrong it should be for ( j = 0; j <= i-1; j++). And Please Mark It as answer if your problem is solved
ArrayList < Integer > arraylist = new ArrayList < Integer > ();
arraylist.add(10010);
arraylist.add(5);
arraylist.add(4);
arraylist.add(2);
for (int i = 0; i < arraylist.size(); i++) {
for (int j = arraylist.size() - 1; j > i; j--) {
if (arraylist.get(i) > arraylist.get(j)) {
int tmp = arraylist.get(i);
arraylist.get(i) = arraylist.get(i);
arraylist.get(j) = tmp;
}
}
}
for (int i: arraylist) {
System.out.println(i);
}
It is giving error while swapping, The LHS should be variable. I understand it.
Set method works here but I do not want to use.
Is there a way to do it without using set method?
Help is really appreciated.
arraylist.get(i)= arraylist.get(i);
arraylist.get(j) =tmp;
You can't assign a value to a method call. As the compiler told you, the left hand side of an assignment must be a variable.
Use set method :
arraylist.set(i,arraylist.get(j));
arraylist.set(j,tmp);
Is there a way to do it without using set method?
No. Unless you wish to convert your ArrayList to an array, sort the array, and update the ArrayList with the sorted array.
Integer[] list={3,5,100,8,17,19};
for(int i=0;i<list.length;i++){
for(int j=i+1;j<list.length;j++){
Integer tempI=list[i];
Integer tempJ=list[j];
if(tempI>tempJ){
list[i]=tempJ;
list[j]= tempI;
}
}
}
for(Integer a:list){
System.out.println(""+a);
}
You can't directly assign value to ArrayList Index. You have to use the set() method.
change your code as follow :
ArrayList < Integer > arraylist = new ArrayList < Integer > ();
arraylist.add(10010);
arraylist.add(5);
arraylist.add(4);
arraylist.add(2);
for (int i = 0; i < arraylist.size(); i++) {
for (int j = arraylist.size() - 1; j > i; j--) {
if (arraylist.get(i) > arraylist.get(j)) {
int tmp = arraylist.get(i);
arraylist.set(i,arraylist.get(j)) ;
arraylist.set(j,tmp);
}
}
}
for (int i: arraylist) {
System.out.println(i);
}
The shortest way to swap to list values is the following:
arraylist.set(i, arraylist.set(j, arraylist.get(i)));
You can utilize the fact that set method returns the previous value. This algorithm is implemented in Collections.swap standard method, so you can use it:
Collections.swap(arraylist, i, j);
Problem in swapping:
int tmp = arraylist.get(i);
arraylist.set(i,arraylist.get(j)) ;
arraylist.set(j,tmp);
import java.util.ArrayList;
import java.util.Scanner;
class Main{
public static String titleCase(String s){
String is = s;
String ret = "";
ret += is.substring(0, 1).toUpperCase();
ret += is.substring(1).toLowerCase();
return ret;
}
public static ArrayList<String> sort(ArrayList<String> list){
for(int z = 0; z <list.size()-1; z++){
if(list.get(z).compareTo(list.get(z+1)) >0){
list.add(z, list.get(z+1));
list.remove(z+2);
}
}
return list;
}
public static void main(String[] args) {
Scanner scan = new Scanner (System.in);
ArrayList<String> names = new ArrayList();
int x = 0;
int count = 0;
String output = "";
while(x != 1){
System.out.println("Enter the next name:");
String temp = scan.next();
temp = titleCase(temp);
if(temp.toLowerCase().equals("stop")){
x = 1;
}
else{
names.add(count, temp);
count++;
}
}
names.equals(sort(names));
System.out.println(names.toString());
}
}
`import java.util.ArrayList;
import java.util.Collections;
public class SortList {
public static void main(String[] args){
ArrayList<Integer> listToSort = new ArrayList<Integer>();
listToSort.add(10010);
listToSort.add(5);
listToSort.add(4);
listToSort.add(2);
System.out.println("List to SOrt"+listToSort);
for(int i= 0;i<listToSort.size();i++)
{
for (int j=0;j<listToSort.size();j++){
if(listToSort.get(j)>listToSort.get(i)){
int temp=listToSort.get(i);
int temp2=listToSort.get(j);
listToSort.set(j,temp);
listToSort.set(i,temp2);
}
}
}
System.out.println("List SOrted"+listToSort);
}
}
Try using this:
public static void SortingArrayListAsc(List<Integer> list) {
for (int i = 0; i < list.size(); i++) {
for (int j = 0; j < list.size(); j++) {
if (list.get(i) < list.get(j)) {
Integer temp = list.get(i);
list.set(i, list.get(j));
list.set(j, temp);
}
}
}
}
public int [] sortArrayByType(int[] array,String sortType){
for(int i=0;i<array.length;i++){
for(int j=i+1;j<array.length;j++){
if(sortType.equals("asc")) {
if (array[i] > array[j]) {
swapValues(array, i, j);
}
}else{
if (array[i] < array[j]) {
swapValues(array, i, j);
}
}
}
}
return array;
}
private void swapValues(int[] array, int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
public class SortUsingCollection{
public static void main(String[] args){
ArrayList al=new ArrayList();
al.add(32);
al.add(3);
al.add(23);
al.add(80);
al.add(38);
al.add(31);
al.add(90);
al.add(8);
ListIterator i=al.listIterator();
int k=0;
while(i.hasNext()){
int loop=0;
Integer n=(Integer)i.next();
ListIterator j=al.listIterator();
while(loop<k){
if(j.hasNext())
j.next();
loop++;
}
while(j.hasNext()){
Integer m=(Integer)j.next();
if(m<n){
j.set(n);
i.set(m);
n=m;
}
}
k++;
}
System.out.println(al);
}
}
I have the below program for sorting Strings based on length. I want to print the shortest element first. I don't want to use Comparator or any API to do this. Where I am going wrong?
public class SortArrayElements {
public static void main(String[] args) {
String[] arr = new String[]{"Fan","dexter","abc","fruit","apple","banana"};
String[] sortedArr = new String[arr.length];
for(int i=0;i<sortedArr.length;i++)
{
sortedArr[i] = compareArrayElements(arr);
}
System.out.println("The strings in the sorted order of length are: ");
for(String sortedArray:sortedArr)
{
System.out.println(sortedArray);
}
}
public static String compareArrayElements(String[] arr) {
String temp = null;
for(int i=0;i<arr.length-1;i++)
{
temp = new String();
if(arr[i].length() > arr[i+1].length())
temp = arr[i+1];
else
temp = arr[i];
}
return temp;
}
}
If you really want to learn Java: use a Comparator. Any other way is bad Java code.
You can however rewrite the Comparator system if you want, it will teach you about proper code structuring.
For your actual code, here are some hints:
Using the proper algorithm is much more important than the Language you use to code. Good algorithms are always the same, no matter the language.
Do never do new in loops, unless you actually need to create new objects. The GC says "thanks".
Change the compareArrayElements function to accept a minimum size and have it return the smallest String with at least minimum size.
You could cut out those Strings that you have considered to be the smallest (set them to null), this will however modify the original array.
Use bubble sort, but instead of comparing ints, just compare String lengths.
I won't write the code for you. You will have to do a little bit of research on this algorithm. Google is your best friend as a programmer.
Good luck.
References:
Bubble sort in Java
Sorting an array of strings
Implement bubbleSort() and swap(). My implementations mutate the original array, but you can modify them to make a copy if you want.
public class SortArrayElements {
public static void main(String[] args) {
String[] arr = new String[]{"Fan", "dexter", "abc", "fruit", "apple", "banana"};
bubbleSort(arr);
System.out.println("The strings in the sorted order of length are: ");
for (String item : arr) {
System.out.println(item);
}
}
// Mutates the original array
public static void bubbleSort(String[] arr) {
boolean swapped = false;
do {
swapped = false;
for (int i = 0; i < arr.length - 1; i += 1) {
if (arr[i].length() > arr[i + 1].length()) {
swap(arr, i, i + 1);
swapped = true;
}
}
} while (swapped);
}
// Mutates the original array
public static void swap(String[] arr, int index0, int index1) {
String temp = arr[index0];
arr[index0] = arr[index1];
arr[index1] = temp;
}
}
Okay, there is the code completely based on loops and on bubble sort. No sets are there as you wanted it. This is a pure loop program so you could understand the nested loops, plus it doesn't change the index or something of the string
import java.util.*;
class strings {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
ArrayList<String> a = new ArrayList<String>(2);
System.out.println("Start entering your words or sentences.");
System.out.println("Type stop to stop.");
String b;
int c = 0, d;
do {
b = in.nextLine();
b = b.trim();
a.add(b);
c++;
}
while (!b.equalsIgnoreCase("stop"));
if (c > 1)
a.remove(a.size() - 1);
System.out.println("Choose the sort you want. Type the corresponding
number");
System.out.println("1. Ascending");
System.out.println("2. Descending");
int sc=in.nextInt();
switch(sc) {
case 1: {
int sag[] = new int[a.size()];
for (int jk = 0; jk < a.size(); jk++) {
b = a.get(jk);
c = b.length();
sag[jk] = c;
}
int temp;
for (int i = 0; i < a.size() - 1; i++) {
for (int j = 0; j < a.size() - 1; j++) {
if (sag[j] > sag[j + 1]) {
temp = sag[j + 1];
sag[j + 1] = sag[j];
sag[j] = temp;
}
}
}
ArrayList saga = new ArrayList();
for (int i = 0; i < sag.length; i++) {
saga.add(sag[i]);
}
for (int i = 0; i < saga.size(); i++) {
for (int j = i + 1; j < saga.size(); j++) {
if (saga.get(i).equals(saga.get(j))) {
saga.remove(j);
j--;
}
}
}
for (int i = 0; i < saga.size(); i++) {
for (int j = 0; j < a.size(); j++) {
String jl = a.get(j);
if (saga.get(i).equals(jl.length()))
System.out.println(jl);
}
}
break;
}
case 2: {
int sag[] = new int[a.size()];
for (int jk = 0; jk < a.size(); jk++) {
b = a.get(jk);
c = b.length();
sag[jk] = c;
}
int temp;
for (int i = 0; i < a.size() - 1; i++) {
for (int j = 0; j < a.size() - 1; j++) {
if (sag[j] < sag[j + 1]) {
temp = sag[j + 1];
sag[j + 1] = sag[j];
sag[j] = temp;
}
}
}
ArrayList saga = new ArrayList();
for (int i = 0; i < sag.length; i++) {
saga.add(sag[i]);
}
for (int i = 0; i < saga.size(); i++) {
for (int j = i + 1; j < saga.size(); j++) {
if (saga.get(i).equals(saga.get(j))) {
saga.remove(j);
j--;
}
}
}
for (int i = 0; i < saga.size(); i++) {
for (int j = 0; j < a.size(); j++) {
String jl = a.get(j);
if (saga.get(i).equals(jl.length()))
System.out.println(jl);
}
}
break;
}
}
}
}
For instance, the following:
ArrayList<String> str = new ArrayList<>(
Arrays.asList(
"Long", "Short", "VeryLong", "S")
);
By lambda:
str.sort((String s1, String s2) -> s1.length() - s2.length());
By static Collections.sort
import static java.util.Collections.sort;
sort(str, new Comparator<String>{
#Override
public int compare(String s1, String s2) {
return s1.lenght() - s2.lenght()
}
});
Both options are implemented by default sort method from List interface
Let's take a following array of String inputArray = ["abc","","aaa","a","zz"]
we can use Comparator for sorting the given string array to sort it based on length with the following code:
String[] sortByLength(String[] inputArray) {
Arrays.sort(inputArray, new Comparator<String>(){
public int compare(String s1, String s2){
return s1.length() - s2.length();
}
});
return inputArray;
}
//sort String array based on length
public class FirstNonRepeatedString {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Please Enter your String");
String str = in.nextLine();
String arrString[] = str.split("\\s");
arrString = sortArray(arrString);
System.out.println("Sort String ");
for(String s:arrString){
System.out.println(s);
}
}
private static String[] sortArray(String[] arrString) {
int length = arrString.length;
String s;
for (int i = 0; i < length ; i++) {
s= new String();
for(int j = 0; j < length; j++ ){
if(arrString[i].length()< arrString[j].length()){
s = arrString[i];
arrString[i] = arrString[j];
arrString[j] = s;
}
}
}
return arrString;
}
}
import java.util.*;
public class SortStringBasedOnTheirLength {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
System.out.println("Enter String:");
String str=sc.nextLine();
String[] str1=str.split("\\s");
for(int i=0;i<str1.length;i++)
{
for(int j=i+1;j<str1.length;j++)
{
if(str1[i].length()>str1[j].length())
{
String temp= str1[i];
str1[i]=str1[j];
str1[j]=temp;
}
}
}
for(int i=0;i<str1.length;i++)
{
System.out.print(str1[i]+" ");
}
}
}