Compare String elements in an Array-CodePudding

I am trying to compare elements in Array. for example,

labels = ["abc1","abc2","abc3","abc4"]

I want to take the String with the highest values. In this case its abc4. I'm pretty new to coding so if anyone could help me with the logic, it would be great.

CodePudding user response：

Your question need improvement, based on what you said, lets remove all the abc from the Strings ,get the max integer and then return or print "abc" concatenated to the max number :

import java.util.Arrays;
import java.util.stream.IntStream;

public class Solution {

    public static void main(String[] args) throws Throwable {
        int numberOfElements = 100;
        String[] labels = new String[numberOfElements];

        Arrays.setAll(labels, element -> "abc"   element);

        int max = Arrays.stream(labels).mapToInt(element -> Integer.parseInt(element.substring(3))).max().getAsInt();

        System.out.println(String.join(" | ", labels));

        System.out.println();
        System.out.println();
        System.out.println("The max here is : ");
        System.out.println("abc"   max);
    }
}

Output here :

abc0  |  | abc1  |  | abc2  |  | abc3  |  | abc4   ...... || abc99
    
The max here is : 
    abc99

CodePudding user response：

What you are looking for is a method that compares each string to each other.

Java has a built-in method way to do this for the String class, called the compareTo method. As it's name suggests, it compares one string to another.

String a = "abc1";
String b = "abc2";
String c = "abc1";
System.out.println(a.compareTo(b)); // prints a negative number because `a` is smaller than `b`
System.out.println(b.compareTo(a)); // prints a positive number because `b` is bigger than `a`
System.out.println(c.compareTo(a)); // prints 0 because `a` and `b` have the same letters.

See the official java doc for the compareTo method:

[Returns]the value 0 if the argument string is equal to this string; a value less than 0 if this string is lexicographically less than the string argument; and a value greater than 0 if this string is lexicographically greater than the string argument.

The way you could use this in your example would be:

String biggest = labels[0];
for(int i = 1; i < labels.length; i  ){
  if(biggest.compareTo(labels[i]) < 0) biggest = labels[i];
}
System.out.println(biggest);

Note: For more details on how this method chooses which one is "bigger", see the java doc (linked above). If you have your own rules about which one should be bigger, then you can make your own method to define that.

UPDATE: For example, see XtremeBaumer's comment

"abc20".compareTo("abc100") = 1. Indicating that abc20 is bigger than abc100, thus making compareTo() not necessarily useful for the task

CodePudding user response：

Try something like this:

    var strings = new ArrayList<String>();
    strings.add("abc1");
    strings.add("abc2");
    strings.add("abc3");
    strings.add("abc4");


    var integers = strings
            .stream()
            .map(string -> Integer.valueOf(string.replaceAll("[^0-9] ", "")))
            .collect(Collectors.toList());
    var max = Collections.max(integers);
    var indexMax = integers.indexOf(max);
    var maxString = strings.get(indexMax);
    System.out.println(maxString);