I have the following array
String[] arrKey = new String[] {"A","B","C","D","E",......"Y","Z"};
I would like to search the array and give me the index of where the letter is. Say for example I want to search for the letter "E" and when I search the array it should give me the position of "E" so I should get index position 4. I don't want to do this in a loop. Is it possible? I have been looking all around and can't find an answer.
I don't want to do this in a loop.
There has to be a loop somewhere - either in your code or library code.
So yes, you can use
int index = Arrays.asList(arrKey).indexOf("E");
... but that will loop under the covers.
If you know that your array is sorted to start with, you can use:
int index = Arrays.binarySearch(arrKey, "E");
That will be more efficient - but it's still a loop...
Of course, if you know that your array is always A-Z, then you can do it without a loop - but I assume your real case is more generalized...
int index = Arrays.asList(arrKey).indexOf("E"); I get -1 as my answer regardless of which one letter I use. When I try this one : int index = Arrays.binarySearch(arrKey, "E"); I get this error: Exception in thread "main" java.lang.ClassCastException: java.lang.Character cannot be cast to java.lang.String I assume it has to do with the type of array?
String c = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"; then char[] arrKey = c.toCharArray(); and your code int index = Arrays.binarySearch(arrKey, "E"); worked for me. I am not sure how it works but from what you said it loops? I am just curious as I am learning JAVA as when you look at the line there is no loop there just a line so I presume that JAVA (in the background) does it's own loop for that statement! Is that right?String[]). Your code should work if you use 'E' instead of "E". And yes, all of these loop. But if you've really got A...Z, you can just use 'E' - 'A'... It would really help if you'd give us an accurate impression of what you're trying to achieve.It is basically a interview question, to test your recursion skills, here is how you do it.
private static <T> int findIndex(T[] items, T item, int index) {
if (items.length == index) {
return -1;
}
if (items[index].equals(item)) {
return index;
}
return findIndex(items, item, index+1);
}
Here is how you can run it
int val = findIndex(new String[]{"A","B","C"}, "C", 0);
If you have to deal with chars and not strings have a look at the getNumericValue method.
If you want a more general solution you should consider using a Map<String,Integer> rather than an array.
Use indexOf
return arrKey.get(arrKey.indexOf("E"));
It only possible if you assume you have this array (or one similar) as you can calculate the index.
String s = "E";
int index = s.charAt(0) - 'E'; // == 4
Here is the code I am working on. This is what I have at the moment. It is just a practice round for me to practice JAVA and understand the Ceaser Cipher.
public class CeaserCipher
{
public static void main (String [] args) {
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter test letters for Caesar cipher in capitals");
String input = keyboard.nextLine();
char[] strArray = input.toCharArray();
System.out.print("What is the key: ");
int key = keyboard.nextInt();
//String[] arrKey = new String[] {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
String c = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char[] arrKey = c.toCharArray();
for (int i = 0; i < strArray.length ; i++){
char cipherValue = strArray[i];
int index = Arrays.binarySearch(arrKey, cipherValue);
int j = (key + index)%26;
System.out.print(arrKey[j]);
}
}
}
char instead of String?'E' in a char[], or "E" in a String[].You can use Recursion for this case:
let arr = ["A","B","C","D","E","Y","Z"];
let index = 0;
function find(str){
if(str[0] === "E")
{
return str[0];
}
else {
str.shift();
//removing 1st element from the str array and calling the function again
index++;
return find(str);
}
}
console.log("Found",find(arr));
console.log("at Index:",index);
console.log("Peace Out! XD");
