Converting String to char? - java

I need to take a String and then work out how many times each letter appears in the string. I thought of maybe converting each letter of the string into an individual char. Is there any way to do this?

you can use this
myString.toCharArray();
to get a char[] out of it. Or just use this :
myString.charAt(i);

Checkout
String#toCharArray() method in String API. This method returns a char[] array with all characters in the String. You can get each character from the array using the index. say, char[i] to return the character of the string in index 'i'. Please note that Array index starts from '0'

Use yourStringVariable.toCharArray()
char[] array = yourStringVariable.toCharArray();

Map<Character, Integer> counts = new HashMap<>();
for (int i=0;i<str.length();i++) {
Integer count = counts.get(str.charAt(i));
if (count == null) {
count = 1;
} else {
count = count+1;
}
counts.put(str.charAt(i), count);
}
The end result is a map containing all of the characters within the string and a count of how many of each one was found.

Related

How to replace first and middle char in string

I need to replace first and middle char in string but without builder and etc, just with replace but idk how to make it.
String char = JOptionPane.showInputDialog(null, "Input string with more than 3 char");
if (char.length() < 3) {
JOptionPane.showMessageDialog(null, "Wrong input");
I just made this code and that is it, idk how to continue.
Example: input - pniut
I tried with smth like char.length / 2 but cant.
You can convert your string to a character array, and then swap the characters at 0 and middle position. Then convert the array back to String. e.g. I hard coded 2 here but like you mentioned in comments, you will need to figure out the character at the middle position.
String str = "input";
int mid = -1;
if(str.length() % 2 == 0) {
str.length() / 2 - 1
} else {
str.length() / 2;
}
char[] arr = str.toCharArray();
char temp = '0';
temp = arr[0];
arr[0] = arr[mid];
arr[mid] = temp;
String.valueOf(arr);
The value of the middle character, you will need to find out, like you said in the comments.
Since String objects are immutable, converting the original String to a char[] via toCharArray(), replace the characters, then making a new String from char[] via the String(char[]) constructor would work as shown below:
char[] c = character.toCharArray();
// Change characters at desired indicies
c[0] = 'p'; // first character
c[character.length()/2] = 'i'; // approximate middle character
String newString = new String(c);
System.out.println(newString); // "pniut"
Simple answer: not possible (for generic cases).
Meaning: all variants of String.replace() work by replacing one thing with another. There is no notion of using an index anywhere. So you can't say "replace index 1 with A" and "index 3 with B".
The simply solution is to push the string into a char[], to then swap/replace individual characters via index.
I'm betting the goal of the lesson is to learn how to use the API. So would start here Java API. Go to java.lang.String.
I would focus on the .toCharArray() method and the constructor that takes a char[] as an argument. You need to do this because a String is immutable, and cannot be changed. A char[], however can be altered, allowing you to modify the first and middle slots. You can then take your altered array and convert it back into a String.

find how many numberof times a character from a String is repeated in java

I have a String of say, length is 5000. I want to find out the number of times the letter 'R' (Case Sensitive) is used. Below are the two passable solution...
Convert to char array, loop it to perform a condition to check and increment a counter.
Use Substring() with the character 'R' to get an array which could fetch a array. So, the total length of the array +1 will be number of times, the 'R' character in the string(This is not a better solution)
Help me out with the efficient solution on the cards for this. Thanks.
try this one
String text = "ABCabcRRRRRrrr";
int count = text.length() - text.replace("R", "").length();
If you're using java >= 8 you can use the Streams:
public static void main(String args[]) {
String str= "abcderfgtretRetRotpabcderfgtretRetRotp"
System.out.println(str.chars().filter(c -> c == 'R').count());
}
String str = //the actual string
for(int i=0;i<str.length();++i)
{
if(str.charAt(i)=='R')
{
capitalRCount++;
}
}

Java Get first character values for a string

I have inputs like
AS23456SDE
MFD324FR
I need to get First Character values like
AS, MFD
There should no first two or first 3 characters input can be changed. Need to get first characters before a number.
Thank you.
Edit : This is what I have tried.
public static String getPrefix(String serial) {
StringBuilder prefix = new StringBuilder();
for(char c : serial.toCharArray()){
if(Character.isDigit(c)){
break;
}
else{
prefix.append(c);
}
}
return prefix.toString();
}
Here is a nice one line solution. It uses a regex to match the first non numeric characters in the string, and then replaces the input string with this match.
public String getFirstLetters(String input) {
return new String("A" + input).replaceAll("^([^\\d]+)(.*)$", "$1")
.substring(1);
}
System.out.println(getFirstLetters("AS23456SDE"));
System.out.println(getFirstLetters("1AS123"));
Output:
AS
(empty)
A simple solution could be like this:
public static void main (String[]args) {
String str = "MFD324FR";
char[] characters = str.toCharArray();
for(char c : characters){
if(Character.isDigit(c))
break;
else
System.out.print(c);
}
}
Use the following function to get required output
public String getFirstChars(String str){
int zeroAscii = '0'; int nineAscii = '9';
String result = "";
for (int i=0; i< str.lenght(); i++){
int ascii = str.toCharArray()[i];
if(ascii >= zeroAscii && ascii <= nineAscii){
result = result + str.toCharArray()[i];
}else{
return result;
}
}
return str;
}
pass your string as argument
I think this can be done by a simple regex which matches digits and java's string split function. This Regex based approach will be more efficient than the methods using more complicated regexs.
Something as below will work
String inp = "ABC345.";
String beginningChars = inp.split("[\\d]+",2)[0];
System.out.println(beginningChars); // only if you want to print.
The regex I used "[\\d]+" is escaped for java already.
What it does?
It matches one or more digits (d). d matches digits of any language in unicode, (so it matches japanese and arabian numbers as well)
What does String beginningChars = inp.split("[\\d]+",2)[0] do?
It applies this regex and separates the string into string arrays where ever a match is found. The [0] at the end selects the first result from that array, since you wanted the starting chars.
What is the second parameter to .split(regex,int) which I supplied as 2?
This is the Limit parameter. This means that the regex will be applied on the string till 1 match is found. Once 1 match is found the string is not processed anymore.
From the Strings javadoc page:
The limit parameter controls the number of times the pattern is applied and therefore affects the length of the resulting array. If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter. If n is non-positive then the pattern will be applied as many times as possible and the array can have any length. If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.
This will be efficient if your string is huge.
Possible other regex if you want to split only on english numerals
"[0-9]+"
public static void main(String[] args) {
String testString = "MFD324FR";
int index = 0;
for (Character i : testString.toCharArray()) {
if (Character.isDigit(i))
break;
index++;
}
System.out.println(testString.substring(0, index));
}
this prints the first 'n' characters before it encounters a digit (i.e. integer).

Read a string and 2 index variables then swap those characters in specified index

My program needs to read a string and two index values from the user, then swap the characters with the specified index values, and save the resulting string in a variable.
Right now all i can do is swap the 2 first letters of the user input string and i am unable to figure out how to swap 2 index values from the user
Use String.toCharArray() to convert the input to a char[]. Then you can work with the indices and swap the required characters. Then all you need is to construct a new string from the array. Refer to the String javadocs.
You could swap the characters like this assuming you have the original String and the two indices.
public static String swapChars(String word, int firstIndex, int secondIndex) {
if (Math.min(firstIndex, secondIndex) < 0 || Math.max(firstIndex, secondIndex) > word.length() - 1) { // If either of the indices are negative or too large, throw an exception
throw new IllegalArgumentException("Indices out of bounds!");
}
if (firstIndex == secondIndex) { // If they are equal we can return the original string/word
return word;
}
char[] characters = word.toCharArray(); // Make a char array from the string
char first = characters[firstIndex]; // Store the character at the first index
characters[firstIndex] = characters[secondIndex]; // Change the character at the first index
characters[secondIndex] = first; // Change the character at the second index using the stored character
return new String(characters); // Return a newly built string with the char array
}

Making multiple searches in a string using scanner or string methods (java)

I need to extract several integers from a string that looks like this:
22:43:12:45
I need to extract 22, 43, 12, and 45 as separate integers. I need to use string methods or scanner methods to read up until : and extract the 22. Then read between the first : and second : to give me 43, and so on.
I can extract up to the first : no problem, but I down know what to do thereafter.
Any help would be much appreciated.
String[] parts = str.split(":");
int[] numbers = new int[parts.length];
Iterate over this String array to get an array of integers:
int index = 0;
for(String part : parts)
numbers[index++] = Integer.parseInt(part);
You should look at String.split method . Given a regular expression, this methods splits the string based on that. In your case the regular expression is a ":"
String s ="22:43:12:45";
int i =0;
for (String s1 : s.split(":")) { // For each string in the resulting split array
i = Integer.parseInt(s1);
System.out.println(i);
}
The split returns a string array with your string separated. So in this case , The resulting string array will have "22" on the 0th position, "43" on the first position and so on. To convert these to integers you can use the parseInt method which takes in a string and gives the int equivalent.
You can use only indexOf and substring methods of String:
String text = "22:43:12:45";
int start = 0;
int colon;
while (start < text.length()) {
colon = text.indexOf(':', start);
if (colon < 0) {
colon = text.length();
}
// You can store the returned value somewhere, in a list for example
Integer.parseInt(text.substring(start, colon)));
start = colon + 1;
};
Using Scanner is even simpler:
String text = "22:43:12:45";
Scanner scanner = new Scanner(text);
scanner.useDelimiter(":");
while (scanner.hasNext()) {
// Store the returned value somewhere to use it
scanner.nextInt();
}
However, String.split is the shortest solution.
Make a regex like
(\d\d):(\d\d):(\d\d):(\d\d)

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