I want to take the value of a ASCII value(Saved as a string) and convert it to the character to reveal a message. I tried this and it keeps throwing an index out of bound at the declaration of the int b.It also shows that str and b do not have a value
String value = "104 101 108 108 111";
char[] ch = new char[value.length()];
for (int i = 0; i < value.length(); i++) {
ch[i] = value.charAt(i);
}
System.out.println(ch.length);
String ans = "";
int i = 0;
while (i+2 < ch.length) {
int b= ch[i]+ch[i++]+ch[i+2];
String str = new Character((char) b).toString();
System.out.println(str);
System.out.println(b);
ans = ans+str;
i=i+3;
}
Using string split function
String value = "104 101 108 108 111";
String[] arrOfStr = value.split(" ");
String ans = "";
for(String str : arrOfStr) {
String str1 = Character.toString((char)Integer.parseInt(str));
ans += str1;
}
System.out.println(ans); // output: hello
We can switch the Imperative code to Declarative code using Java 8 Streams.
Key points to observe:
Declarative style is more readable and easy to write.
String Joiner is faster than simple String Concatenation.
No need to write an iterator.
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
String value = "104 101 108 108 111";
Arrays.stream(value.split(" ")) // Starting a stream of String[]
.mapToInt(Integer::parseInt) // mapping String to int
.mapToObj(Character::toChars) // finding ASCII char from int
.forEach(System.out::print); // printing each character
}
}
If you wish to store the result and then print it, this is how is done.
import java.util.Arrays;
import java.util.stream.Collectors;
public class Main {
public static void main(String[] args) {
String value = "104 101 108 108 111";
String result = Arrays.stream(value.split(" ")) // Starting a stream of String[]
.mapToInt(Integer::parseInt) // mapping String to int
.mapToObj(Character::toChars) // finding ASCII char from int
.map(String::new) // convert char to String
.collect(Collectors.joining()); // combining individual result using String Joiner
System.out.println(result);
}
}
Comments to code:
There is a built-in method for for getting a char[] with the characters of a string, so the following two blocks of code are the same:
// Code from question
char[] ch = new char[value.length()];
for (int i = 0; i < value.length(); i++) {
ch[i] = value.charAt(i);
}
// Using built-in method
char[] ch = value.toCharArray();
It is better to use a for loop when increment a value while looping. The following two ways of writing the loop behave the same, but the for loop keeps the loop logic together:
// Code from question
int i = 0;
while (i+2 < ch.length) {
// some code here
i=i+3;
}
// Using for loop
for (int i = 0; i + 2 < ch.length; i=i+3) {
// some code here
}
The following line of code is entirely wrong:
int b= ch[i]+ch[i++]+ch[i+2];
i++ increments the value is i, but it is the value before the increment that is used in the expression, which means that if i = 0 before the line, the result is the same as this code:
int b = ch[0] + ch[0] + ch[2];
i = i + 1;
You need to replace i++ with i + 1, and realize that those are not the same.
Since you no longer increment the value of i by 1 in that statement, the loop much be changed from i=i+3 to i = i + 4, to correctly skip the spaces in the input.
The value of ch[i] is a char value, which is widened to an int value by the use of the + operator. The int value of a char is the Unicode Code Point value, which for your text is also the same as the ASCII code for the character.
This means that if i = 0, the expression would (after fixing issue #1) evaluate as:
int b = ch[0] + ch[1] + ch[2];
int b = `1` + `0` + `4`;
int b = 49 + 48 + 52;
int b = 149;
That matches the output from running the code is in question, where the second printed number is 149 (after fixing issue #1).
What you really wanted was to get the substring "104" and convert that to a number, then cast that ASCII code value to a char, like this:
String numberStr = value.substring(i, i + 3); // E.g. "104"
int number = Integer.parseInt(numberStr); // E.g. 104
String str = String.valueOf((char) number); // E.g. "h"
With that, you no longer need the char[], so the final code would be:
String value = "104 101 108 108 111";
String ans = "";
for (int i = 0; i + 2 < value.length(); i += 4) {
String numberStr = value.substring(i, i + 3);
int number = Integer.parseInt(numberStr);
String str = String.valueOf((char) number);
ans = ans + str;
}
System.out.println(ans);
Output
hello
Related
Hi guys I am busy with breaking / splitting Strings.
However the String is not fixed so when the input changes the program still has to work with any character input.
Till now I got this far but I got lost.
I have made an array of characters and set the size of the array equal to the lenght of any string that is will get as input. I made a for loop to loop through the characters of a string.
how do I insert my string now into the array because I know that my string is not yet in there? Then when its finally looping through the characters of my string is has to printout numbers and operands on different lines. So the ouput would look like in this case like this;
1
+
3
,
432
.
123
etc
I want to do this without using matchers,scanner, etc. I want to use basic Java techniques like you learn in the first 3 chapters of HeadfirstJava.
public class CharAtExample {
public static void main(String[] args) {
// This is the string we are going to break down
String inputString = "1+3,432.123*4535-24.4";
int stringLength = inputString.length();
char[] destArray = new char[stringLength];{
for (int i=0; i<stringLength; i++);
}
You could use Character.isDigit(char) to distinguish numeric and not numeric chars as actually this is the single criteria to group multiple chars in a same line.
It would give :
public static void main(String[] args) {
String inputString = "1+3,432.123*4535-24.4";
String currentSequence = "";
for (int i = 0; i < inputString.length(); i++) {
char currentChar = inputString.charAt(i);
if (Character.isDigit(currentChar)) {
currentSequence += currentChar;
continue;
}
System.out.println(currentSequence);
System.out.println(currentChar);
currentSequence = "";
}
// print the current sequence that is a number if not printed yet
if (!currentSequence.equals("")) {
System.out.println(currentSequence);
}
}
Character.isDigit() relies on unicode category.
You could code it yourself such as :
if (Character.getType(currentChar) == Character.DECIMAL_DIGIT_NUMBER) {...}
Or you could code it still at a lower level by checking that the int value of the char is included in the range of ASCII decimal values for numbers:
if(currentChar >= 48 && currentChar <= 57 ) {
It outputs what you want :
1
+
3
,
432
.
123
*
4535
-
24
.
4
It's easier than you might think.
First: to get an array with the chars of your string you just use the toCharArray() method that all strings have. ex. myString.toCharArray()
Second: When you see that a character is not a number, you want to move to the next line, print the character and then move to the next line again. The following code does exactly that :
public class JavaApplication255 {
public static void main(String[] args) {
String inputString = "1+3,432.123*4535-24.4";
char[] destArray = inputString.toCharArray();
for (int i = 0 ; i < destArray.length ; i++){
char c = destArray[i];
if (isBreakCharacter(c)){
System.out.println("\n" + c);
} else {
System.out.print(c);
}
}
}
public static boolean isBreakCharacter(char c){
return c == '+' || c == '*' || c == '-' || c == '.' || c == ',' ;
}
char[] charArray = inputString.toCharArray();
Here is a possible solution where we go character by character and either add to an existing string which will be our numbers or it adds the string to the array, clears the current number and then adds the special characters. Finally we loop through the array as many times as we find a number or non-number character. I used the ASCII table to identify a character as a digit, the table will come in handy throughout your programming career. Lastly I changed the array to a String array because a character can't hold a number like "432", only '4' or '3' or '2'.
String inputString = "1+3,432.123*4535-24.4";
int stringLength = inputString.length();
String[] destArray = new String[stringLength];
int destArrayCount = 0;
String currentString = "";
for (int i=0; i<stringLength; i++)
{
//check it's ascii value if its between 0 (48) and 9 (57)
if(inputString.charAt(i) >= 48 && inputString.charAt(i) <= 57 )
{
currentString += inputString.charAt(i);
}
else
{
destArray[destArrayCount++] = currentString;
currentString = "";
//we know we don't have a number at i so its a non-number character, add it
destArray[destArrayCount++] = "" + inputString.charAt(i);
}
}
//add the last remaining number
destArray[destArrayCount++] = currentString;
for(int i = 0; i < destArrayCount; i++)
{
System.out.println("(" + i + "): " + destArray[i]);
}
IMPORTANT - This algorithm will fail if a certain type of String is used. Can you find a String where this algorithm fails? What can you do to to ensure the count is always correct and not sometimes 1 greater than the actual count?
I have started to write a code that would reverse every second set of 4 characters. eg. 1234567890987654. The result would be displayed as ("1234" "8765" "9098" "4567"). This would then be multiplied as in "1234" * "8765" * "9098" * "4567".
this code doesn't work and I would like some help.
My code currently is
public static void main(String[] args) {
String input = "1234567890987654";
System.out.println(result);
public static int reverse(String a) {
String newa = "";
String str = a;
char ch;
String[] array = str.split("(?<=\\G.{2})");
a = "array[]";
for (int i = 0 ; i < a.length() ; i=i+2) {
ch = a.charAt(i);
newa = ch + newa;
//System.out.println(newcardNum);
}
return newa;
}
Thanks in adavance
Let's start by creating a proper test framework (at least a toy testing framework) -
public static void main(String[] args) {
String input = "1234567890987654";
String expected = "1234" + "8765" + "9098" + "4567";
String result = reverse(input);
if (result.equals(expected)) {
System.out.println("result matches expected result");
} else {
System.out.println("result does not match expected result");
}
System.out.printf("input='%s', result = '%s'%n", input, result);
}
Now we can tell when the result we receive matches our expected result, and what the input and output String(s) are.
Next, we need to fix your reversing algorithm. You want to get 4 character then reverse 4 characters, so each iteration is 8 characters. Something like,
public static String reverse(String a) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < a.length(); i += 8) { // <-- increment by 8.
sb.append(a.substring(i, i + 4)); // <-- pass 1,4
// get characters 4 to 8
StringBuilder sb2 = new StringBuilder(a.substring(i + 4, i + 8));
// reverse and append
sb.append(sb2.reverse().toString());
}
return sb.toString();
}
Output is
result matches expected result
input='1234567890987654', result = '1234876590984567'
Edit Based on your comments below (and edited question), add something like this to main
String[] arr = new String[4];
long calc = 1; // <-- int would overflow
for (int count = 0, i = 0; i < result.length(); count++, i += 4) {
arr[count] = result.substring(i, i + 4);
calc *= Integer.parseInt(arr[count]);
}
System.out.printf("Array = %s, multiplication = %d%n", Arrays.toString(arr), calc);
And now the output is
result matches expected result
input='1234567890987654', result = '1234876590984567'
Array = [1234, 8765, 9098, 4567], multiplication = 449411337361660
I've got this bit of code here:
public class Project1 {
public static void main(String args[])
{
Scanner input = new Scanner(System.in);
System.out.println("Input a binary number");
String binary = input.nextLine();
System.out.println(Conversion(binary));
}
public static int Conversion(String binary)
{
StringTokenizer st = new StringTokenizer(binary, " ");
int n = st.countTokens() - 1; // Used as the power number in b^n for conversion
int result = 0;
while(st.hasMoreTokens()){
int binaryInt = Integer.parseInt(st.nextToken());
result += binaryInt * (1 << n);
n--;
}
return result;
}
}
And it works beautifully... if the input has spaces in between the binary numbers. For example, if the input is 1 1 1 1, then it will rightfully return 15. Cool, but how do I change the tokenizer to not require spaces to split each digit? I tried doing StringTokenizer(binary, ""); and StringTokenizer(binary);, but neither properly split each digit into it's own token.
You will notice that, while StringTokenizer obviously only works with Strings, I've converted the Strings into ints inside my conversion method before returning the result.
# Tony I think you tried to add the ascii values of the digits here
for input value 10
for first run
int binaryInt = binary.char(i) ; so binaryInt would get the value = 49(aski value of char '1')
hence result = 49*2 = 98
in second run
result = 98 + 48 *1 = 146
following function will serve your purpose, for input String "1101" , it returns 13
public int getDecimal(String binaryString){
//binaryString = "1101";
int result = 0;
int n = binaryString.length()-1;
for(int i=0;i<binaryString.length();i++)
{
int num = binaryString.charAt(i);
if(num>=48 && num <=57){
result+=(num-48) * Math.pow(2, n) ;
n --;
}
}
return result;
}
}
How about something like this:
// remove everything that is not a digit
String digitsOnly = binary.replaceAll("[^\\d]","");
//iterate over every digit
for(char digit:digitsOnly.toCharArray()) {
int n = Character.getNumericValue(digit);
// do stuff...
}
If you get a string like "10011010", you don't need a tokenizer, you can just iterate over the string and use charAt:
for (int i = 0; i < s.length(); i++) {
char currentDigit = s.charAt(i);
// Do stuff with the digit
}
Well, this is my first time get here.
I'm trying to figure out the correct way to replace number into letter.
In this case, I need two steps.
First, convert letter to number. Second, restore number to word.
Words list: a = 1, b = 2, f = 6 and k = 11.
I have word: "b a f k"
So, for first step, it must be: "2 1 6 11"
Number "2 1 6 11" must be converted to "b a f k".
But, I failed at second step.
Code I've tried:
public class str_number {
public static void main(String[] args){
String word = "b a f k";
String number = word.replace("a", "1").replace("b","2").replace("f","6").replace("k","11");
System.out.println(word);
System.out.println(number);
System.out.println();
String text = number.replace("1", "a").replace("2","b").replace("6","f").replace("11","k");
System.out.println(number);
System.out.println(text);
}
}
Result:
b a f k
2 1 6 11
2 1 6 11
b a f aa
11 must be a word "k", but it's converted to "aa"
What is the right way to fix this?
Or do you have any other ways to convert letter to number and vice versa?
Thank you.
It would be good to write methods for conversion between number and letter format. I would write some code like this and use it generally instead of hard coding replace each time.
public class test {
static ArrayList <String> letter = new ArrayList<String> ();
static ArrayList <String> digit = new ArrayList<String> ();
public static void main(String[] args) {
createTable();
String test="b a f k";
String test1="2 1 6 11";
System.out.println(letterToDigit(test));
System.out.println(digitToLetter(test1));
}
public static void createTable()
{
//Create all your Letter to number Mapping here.
//Add all the letters and digits
letter.add("a");
digit.add("1");
letter.add("b");
digit.add("2");
letter.add("c");
digit.add("3");
letter.add("d");
digit.add("4");
letter.add("e");
digit.add("5");
letter.add("f");
digit.add("6");
letter.add("g");
digit.add("7");
letter.add("h");
digit.add("8");
letter.add("i");
digit.add("9");
letter.add("j");
digit.add("10");
letter.add("k");
digit.add("11");
letter.add("l");
digit.add("12");
letter.add("m");
digit.add("13");
letter.add("n");
digit.add("14");
letter.add("o");
digit.add("14");
letter.add("p");
digit.add("15");
//Carry so on till Z
}
public static String letterToDigit(String input)
{
String[] individual = input.split(" ");
String result="";
for(int i=0;i<individual.length;i++){
if(letter.contains(individual[i])){
result+=Integer.toString(letter.indexOf(individual[i])+1)+ " ";
}
}
return result;
}
public static String digitToLetter(String input)
{
String[] individual = input.split(" ");
String result="";
for(int i=0;i<individual.length;i++){
if(digit.contains(individual[i])){
result+=letter.get(digit.indexOf(individual[i])) + " ";
}
}
return result;
}
}
I would actually not use replace in this case.
A more generic solution would be to simply convert it to a char and subtract the char a from it.
int n = word.charAt(0) - 'a' + 1;
This should return an int with the value you are looking for.
If you want this to be an string you can easily do
String s = Integer.parseInt(word.charAt(0) - 'a' + 1);
And as in your case you are doing a whole string looping through the length of it and changing all would give you the result
String s = "";
for(int i = 0; i < word.length(); i++) {
if(s.charAt(i) != ' ') {
s = s + Integer.toString(word.charAt(i) - 'a' + 1) + " ";
}
}
and then if you want this back to an String with letters instead
String text = "";
int temp = 0;
for(int i = 0; i < s.length(); i++) {
if(s.charAt(i) == ' ') {
text = text + String.valueOf((char) (temp + 'a' - 1));
temp = 0;
} else if {
temp = (temp*10)+Character.getNumericValue(s.charAt(i));
}
}
You can just reverse the replacement:
String text = number.replace("11","k").replace("2","b").replace("6","f").replace("1","a");
Simplest solution IMO.
When adding other numbers, first replace these with two digits, then these with one.
Replace this:
String text = number.replace("1", "a").replace("2","b").replace("6","f").replace("11","k");
By this:
String text = number.replace("11","k").replace("1", "a").replace("2","b").replace("6","f");
Right now, the first replace you're doing: ("1", "a")
is invalidating the last one: ("11","k")
I think you would need to store the number as an array of ints. Otherwise, there is no way of knowing if 11 is aa or k. I would create a Map and then loop over the characters in the String. You could have one map for char-to-int and one for int-to-char.
Map<Character,Integer> charToIntMap = new HashMap<Character,Integer>();
charToIntMap.put('a',1);
charToIntMap.put('b',2);
charToIntMap.put('f',6);
charToIntMap.put('k',11);
Map<Integer,Character> intToCharMap = new HashMap<Integer,Character>();
intToCharMap.put(1,'a');
intToCharMap.put(2,'b');
intToCharMap.put(6,'f');
intToCharMap.put(11,'k');
String testStr = "abfk";
int[] nbrs = new int[testStr.length()];
for(int i = 0; i< testStr.length(); i++ ){
nbrs[i] = charToIntMap.get(testStr.charAt(i));
}
StringBuilder sb = new StringBuilder();
for(int num : nbrs){
sb.append(num);
}
System.out.println(sb.toString());
//Reverse
sb = new StringBuilder();
for(int i=0; i<nbrs.length; i++){
sb.append(intToCharMap.get(nbrs[i]));
}
System.out.println(sb.toString());
This failed because the replace("1", "a") replaced both 1s with a characters. The quickest fix is to perform the replace of all the double-digit numbers first, so there are no more double-digit numbers left when the single-digit numbers get replaced.
String text = number.replace("11","k").replace("1", "a").
replace("2","b").replace("6","f");
The question is to generate the lexicographically greatest string given some string s.
So the aim is to find lexicographically greatest, unique(no repetitions) substring s1 from s.
We say that some subsequence s1 is greater than another subsequence s2 if s1 has more characters than s2 or s1 is lexicographically greater than s2 if equal length.
I/O are as follows:
Input is: babab
output is: ba
Second input is: nlhthgrfdnnlprjtecpdrthigjoqdejsfkasoctjijaoebqlrgaiakfsbljmpibkidjsrtkgrdnqsknbarpabgokbsrfhmeklrle
Second output is:
tsocrpkijgdqnbafhmle
This is what I wrote for my java code but my code fails on the second test case. Also I'm having a hard time understanding why second output isn't tsrqponmlkjihgfedcba.
Can somebody provide suggestions for a fix or even java code?
I think the algorithm has to be more efficient than generating all possible unique strings, sort them and find lexicographically largest one.
To make the question much clearer, if the input is babab, then all the possible unique combinations would be b, a, ba, ab. And the output will be ba because it's the longest and lexicographically greater than ab.
Note: this is not a homework assignment.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class mostBeautiful {
final static int MAX = 1000000;
static String[] permute;
static void permutation(String prefix, String str, int counter) {
int n = str.length();
//System.out.println("n is: "+ n);
if (n == 0) {
permute[counter] = prefix;
} else {
for (int i = 0; i < n; i++) {
//System.out.println("str is: "+ str);
permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1, n), counter++);
}
}
}
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String s = bf.readLine();
char[] unique = new char[26];
int counter = 0;
String answer = "";
//System.out.println("s is: " + s);
int ascii = 0;
final int asciiAVal = 97;
final int asciiZVal = 122;
for (int i = 0; i < s.length(); i++) {
ascii = (int)s.charAt(i);
if (ascii < asciiAVal || ascii > asciiZVal) {
continue;
}
char ch = s.charAt(i);
unique[ch - 'a'] = ch;
}
String result = "";
for (int j = 25; j >= 0; j--) {
result += unique[j];
}
result = result.trim();
System.out.println(result);
int size = result.length() * (result.length() - 1);
permute = new String[size];
permutation("", result, counter);
for (int i = 1; i < size; i++) {
if (permute[i].compareTo(permute[i - 1]) > 0){
answer = permute[i];
} else {
answer = permute[i - 1];
}
}
System.out.println("answer is: " + answer);
}
}
After thinking about this problem in many ways, I have determined a divide-and-conquer algorithm that gets the results right:
Algorithm - Pseudocode
Assuming some input string, S defined as a concatenation of two substrings A + B, we compute the lexicographically greatest string recursively as:
LexMax(S) = Merge(LexMax(A),LexMax(B))
Where
LexMax(S)
{
if Length(S) = 1
return S
else
{
LMA = LexMax(S[0:Length/2])
LMB = LexMax(S[Length/2:end])
return Merge(LMA,LMB)
}
}
Merge(A,B)
{
Sa = A
Sb = B
for n = 0:Length(A)
{
if Sb contains A[n]
{
if A[n+1:end] contains character > A[n]
Remove A[n] from Sa
else
Remove A[n] from Sb
}
}
return Sa + Sb
}
Java Code
Coming soon!
Example
Given an input string
cefcfdabbcfed
Divide it into
cefcfda
bbcfed
Assuming the function works we have:
LexMax("cefcfda") = "efcda"
LexMax("bbcfed") = "bcfed"
Merging works as follows:
e: efcda bcfed
In both substrings, greater value found to right of e in left substring, remove from left
f: fcda bcfed
In both substrings, no greater value in left substring, remove from right
c: fcda bced
In both substrings, greater value found to right of c in left substring, remove from left
d: fda bced
In both substrings, no greater value in left substring, remove from right
a: fda bce
Not in both substrings, do nothing
Final result:
LexMax(cefcfdabbcfed) = fdabce
This is not a direct answer, but doesn't this code meet the requirement as you explained it in the discussion above?
final String x = "saontehusanoethusnaoteusnaoetuh";
final SortedSet<Character> chars =
new TreeSet<Character>(Collections.reverseOrder());
for (char c : x.toCharArray()) chars.add(c);
System.out.println(chars);
Lexicographic order is an order in which words are displayed in alphabetical order using the appearance of letters in the word.It is also know as dictionary order or alphabetical order.For ex:-"Africa" is smaller than "Bangladesh" ,"He" is smaller than "he".
public class LexicographicExample {
public static void main(String a[]) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter the String:-");
String str = sc.nextLine();
System.out.println("Enter the length");
int count = sc.nextInt();
List<String> list = new ArrayList<String>();
for (int i = 0; i < str.length(); i = i + 1) {
if (str.length() - i >= count) {
list.add(str.substring(i, count + i));
}
}
Collections.sort(list);
System.out.println("Smallest subString:-" + list.get(0));
System.out.println("Largest subString:-" + list.get(list.size() - 1));
}
}
For reference ,refer this link http://techno-terminal.blogspot.in/2015/09/java-program-to-find-lexicographically.html
"tsrqponmlkjihgfedcba" is not the answer because it is not a subsequence of the input. The definition of subsequence requires that the characters of the subsequence occur in the original sequence in the same order. For example, "abc" is a subsequence of "apbqcr", while "cba" is not.
As to the solution, I think a simple greedy algorithm would suffice. First, one has to understand that the maximum possible length of the output is the number of unique symbols (say, N) in the input. Since any output shorter than that would not be the greatest one, it has to be exactly N symbols long. The rest of the procedure is simple and at most quadratic in time complexity: one has to go through the input string and at each step pick the lexicographically highest symbol such that the part of the string to the left of it would still contain all the "unused" symbols.
As an example, consider a string "bacb". The first symbol can be 'a' or 'b', since in both cases the remainder contains both of the other letters. 'b' is greater, so we pick it. Now for "acb" we can only pick 'a' and than 'c' according to that condition, so we end up with "bac" for output.
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
class aaa {
public static void main(String args[]) throws Exception {
Scanner scan = new Scanner(System.in);
// int n = scan.nextInt();
String s = scan.next();
HashMap<Character, Node5> map = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
if (!map.containsKey(s.charAt(i))) {
Node5 node = new Node5();
node.nl.add(i);
node.li = i;
map.put(s.charAt(i), node);
} else {
Node5 rn = map.get(s.charAt(i));
rn.nl.add(i);
rn.li = i;
map.put(s.charAt(i), rn);
}
}
String s1 = "";
int index = -1;
for (int i = 25; i >= 0; i--) {
if (map.containsKey((char) (97 + i))) {
if (map.get((char) (97 + i)).li > index) {
for (int j = 0; j < map.get((char) (97 + i)).nl.size(); j++) {
if (map.get((char) (97 + i)).nl.get(j) > index) {
s1 += (char) (97 + i);
index = map.get((char) (97 + i)).nl.get(j);
}
}
}
}
}
System.out.println(s1);
scan.close();
}
}
class Node5 {
int li;
ArrayList<Integer> nl;
public Node5() {
this.nl = new ArrayList<>();
}
}