I was wondering how you would represent letters as integers in java. I am working on a problem where I have to find the mid letter between a two lettered word. For example, I would choose the word 'go' and provide each letter with an assigned integer value to find the midpoint letter. Can anyone help me out with this or just point me in the right direction to go about solving on how to get the midpoint letter between a two letter word?
That is simple
int a = 'a';
int c = 'c';
char mid = (char) ((a + c) / 2);
System.out.println(mid);
prints
b
(int)str.charAt(i) will get you an integer value (the ASCII value). For "regular" letters, this should allow you to do what you want.
str = "GO";
midLetter = Character.toChars(((int)str.charAt(0) + (int)str.charAt(1))/2);
I think I got the brackets to match...
If by "letters" you're referring to the char primitive type, then they are already represented by integers behind the scenes. From the Java Tutorials:
The char data type is a single 16-bit Unicode character. It has a
minimum value of '\u0000' (or 0) and a maximum value of '\uffff' (or
65,535 inclusive).
So you can assign a char literal to an int variable for example:
int g = 'g';
int o = 'o';
That should be enough to get you started.
In java (and most other language) characters are actually represented as numbers. Google for 'ascii table', and you'll find out lowercase a is actually 97.
Assuming you want to index lowercase a as 0, then given arbitrary character from a string, you can subtract it with the 'a' chacater, and you will get the index
String str = ...;
for(int i=0; i<str.length(); i++) {
char c = str.charAt(i);
int cIndex = c - 'a';
// do something with cIndex...
}
If the given letter is of char then convert the type (int)yourword. Then find the midpoint
Character.getNumericValue returns a numeric value for each character. So, each letter has a unique numeric value via the Character class.
This could be the starting point for your ordering, although you might need to consider case, etc.
You should be able to go back and forth from int to char and perform arithmetic on your char values:
char median = 0;
for(char ch=65; ch<91; ch++) {
System.out.println(ch);
median += ch;
}
median = (char)(median/26);
System.out.println("=====");
System.out.println("Median leter in the alphabet: "+median);
Related
I encountered a code which some parts I do not understand. It has something to do with keeping count of letters in a string. I have commented the part that I do not get. I would appreciate any help. Thank you!
I tried looking it up online but none seem to answer my question.
public class test2 {
static int[] inventory;
public static final int ALPHABET = 26;
public static void main(String[] args) {
inventory = new int [ALPHABET];
String dog = "There goes the dog!";
int size = count(dog);
System.out.println(size);
}
private static int count(String data) {
data = data.toLowerCase();
int size = 0;
for (int i = 0; i < data.length(); i++) {
char ch = data.charAt(i);
if (Character.isLetter(ch)) {
size++;
inventory[ch - 'a']++; // this I don't get
}
}
return size;
}
}
In this case, inventory is an array of size 26 which holds number of times each character appears in the string.
The statement you have put the comment on is trying to deduct the ASCII value of a which is 97 from the character in the string. For example, b's ASCII value is 98 and 'b' - 'a' is 1. So, inventory[1] is incremented i.e b appeared in your string once.
System.out.println(0 + 'c'); //ASCII value of 'c'; will print 99
System.out.println(0 + 'a'); //ASCII value of 'a'; will print 97
System.out.println('c' - 'a'); //Difference of ASCII values of characters; will print 99-97=2
In your case
inventory[ch - 'a']++;
ch will be some character.
ch - 'a' will be the distance of that character from 'a'. For example, as shown above, 'c' - 'a' = 2.
inventory[ch - 'a'] will point to the number at index ch - 'a' in the array.
inventory[ch - 'a']++ will increment that value by 1.
As per the above code, you're trying to count the number of each alphabets.
So, for each letter, let's assume you use a single bucket. Which is done by new int [ALPHABET];. So, you have 26 slots from 0 to 25.
Now, when you're counting the letters: Each letter has a value assigned to it (ASCII value). ASCII value of lower case a a is 97. But, in your slot, you want to add a to slot 0. So, what do you do: you subtract 97 i.e. you subtract value of a for each letter.
So now, a is stored in slot 0, b in slot 1 and so on.
so I am trying to set each of the letter in the alphabet to a number like a = 1
b = 2 c =3 and so on.
int char = "a";
int[] num = new int{26};
for (int i = 0; i <num.length; i++){
System.out.print(i);
But after this i got Stuck so if you possible help me out. So when the users input a word like cat it would out put 3-1-20.
You can subtract 'a' from each char and add 1. E.g.
String input = "cat";
for (char c : input.toCharArray()) {
System.out.print(c - 'a' + 1);
}
The code you posted doesn't compile as you can't assign a String to an int and char is a reserved word (name of a primitive type)
int char = "a";
You also mention that you want the output formatted like this "3-1-20". This is one way to achieve that :
String input = "cat";
String[] out = new String[input.length()];
for (int i = 0; i < input.length(); ++i) {
out[i] = Integer.toString(input.charAt(i) - 'a' + 1);
}
System.out.println(String.join("-", out));
Both versions work only for lowercase English letters (a to z)
Assigning a number to a character is called an "encoding". As computers can only handle numbers internally, this happens all the time. Even this text here is encoded (probably into an encoding called "UTF-8") and then the resulting number is stored somewhere.
One very basic encoding is the so called ASCII (American Standard Code for Information Interchange). ASCII already does what you want, it assigns a number to each character, only that the number for "A" is 65 instead 1.
So, how does that help? We can assume that for the character A-z, the numeric value of a char is equal to the ASCII code (it's not true for every character, but for the most basic ones, it's good enough).
And this is why everyone here tells you to subtract 'A' or 'a': Your character is a char, which is a character, but also the numeric value of that character, so you can subtract 'A' (again, a char) and add 1:
'B' - 'A' + 1 = 2
because...
66 (numeric value of 'B') - 65 (numeric value of 'A') + 1 = 2
Actually, char is not ASCII, but UTF-8, but there it starts to get slightly bit more complex, so ASCII will suffice for the moment.
the best way of doing this is to convert the String to a byte[], like this:
char[] buffer = str.toCharArray();
Then each of the characters can be converted to their byte-value (which are constants for a certain encoding), like this:
byte[] b = new byte[buffer.length];
for (int i = 0; i < b.length; i++) {
b[i] = (byte) buffer[i];
}
Now look at the resulting values and subtract/add some value to get the desired results!
I was given an interview question to output the most frequent occurrence of a character.
Given this string "aaxxaabbaa". The character 'a' is the most frequent.
The following code was something that I found searching on the internet. Note: I implemented it with 2 loops which is in-efficient (different topic)
public static char findMostUsedChar(String str){
char maxchar = ' ';
int maxcnt = 0;
// if you are confident that your input will be only ascii, then this array can be size 128.
// Create a character counter
**int[] charcnt = new int[Character.MAX_VALUE + 1];**
for(int i = 0; i < str.length()-1; i++){
char **ch** = str.charAt(i);
// increment this character's cnt and compare it to our max.
if (**charcnt[ch]**++ >= maxcnt) {
maxcnt = charcnt[ch];
maxchar = ch;
}
}
return maxchar;
}
They declared an int array, found the character at a specific index (i.e. 'a') then used as an index.
After tracing the code out on the debugger in eclipse, I still don't understand how using a character to represent an int index works without explicitly casting it or using charVal.getNumericValue()?? Even most of the S.O. char to int topics explicitly cast.
Thanks in advance.
Array access expressions undergo implicit unary numeric promotion, which will widen an expression to int.
The index expression undergoes unary numeric promotion (ยง5.6.1).
The char datatype is widened to int via its Unicode value, e.g. 'A' -> 65.
I tried to look in previous stack overflow questions but could not find any similar.
I want to store the integer value to char variable but the integer value is stored in another variable.
char[] ch1 = (binary1.toString()).toCharArray();
char[] ch2 = binary2.toString().toCharArray();
if (ch1.length >= ch2.length) {
char[] ch = new char[ch1.length];
int j = 0;
int num1;
int num2;
int num3;
for (int i = 0; i < ch1.length; i++) {
if (j == ch2.length - 1)
j = 0;
num1 = Character.getNumericValue(ch1[i]);
num2 = Character.getNumericValue(ch2[j]);
num3 = num1 ^ num2;
ch[i] = (char) num3;
j++;
}
String str = new String(ch);
return str;
}
Here what is happening is that I am getting null values in many cases.
I tried to look in the Character class but could not find any function.
If there is any way then please tell.
Thanks in advance.
EDIT : I need to store either 0 or 1
EDIT : ch1[] & ch2[] ar of char type
EDIT : I got a temporary solution to my problem but for a wider concept how should we type cast integer to char if size is same and when getting null values.
You can do
char ch = num3 == 0 ? '0' : '1';
When you call Character.getNumericValue(someChar); You are actually getting the integer literal of the value.
Lets say someChar='4'; in this case, The above method will give you a value 4.
But you cannot get numaric value if the char has value other than 0 to 9
If you really want the integer value of what so ever byte value stored in the char variable, then you have to type cast it directly to an integer as below.
char x = 'Z';
int z = x;
System.out.println(z);
the above code snippet will print 90 as Z is equalent to 90 in ascii.
Maybe your conversion from
char ch = (char)num3; is not right?
Maybe you can try to use something like
Character.toChars(65)
See this answer for information on that:Converting stream of int's to char's in java
If you know that your integer is a single digit number, then you can use Character's forDigit method to do the conversion from int to char
Character.forDigit(9,10) --> '9'
From the Official docs,
Determines the character representation for a specific digit in the
specified radix.
Parameters:
digit - the number to convert to a character.
radix - the radix.
Returns:
the char representation of the specified digit in the specified radix.
Hello I am trying to write a little segment of code that checks if each char in a char array is lower case or uppercase. Right now it uses the char's ASCII number to check. After it checks it should convert the char to upper case if it is not already so:
for (int counter = 0; counter < charmessage.length; counter++) {
if (91 - charmessage[counter] <= 0 && 160 - charmessage[counter] != 0) {
charmessage[counter] = charmessage[counter].toUpperCase();
}
}
charmessage is already initialized previously in the program. The 160 part is to make sure it doesn't convert a space to uppercase. How do I get the .toUpperCase method to work?
I would do it this way. First check if the character is a letter and if it is lowercase. After this just use the Character.toUpperCase(char ch)
if(Character.isLetter(charmessage[counter]) && Character.isLowerCase(charmessage[counter])){
charmessage[counter] = Character.toUpperCase(charmessage[counter]);
}
You can use the Character#toUpperCase for that. Example:
char a = 'a';
char upperCase = Character.toUpperCase(a);
It has some limitations, though. It's very important you know that the world is aware of many more characters that can fit within the 16-bit range.
String s = "stackoverflow";
int stringLenght = s.length();
char arr[] = s.toCharArray();
for (int i = stringLenght - 1; i >= 0; i--) {
char a = (arr[i]);
char upperCase = Character.toUpperCase(a);
System.out.print(upperCase);
}