sorting of arraray's number element in javascript
let numbers = [0, 1, 2, 3, 10, 20, 30];
numbers.sort((a, b) => a - b);
console.log(numbers);sorting of arraray's number element in javascript
let numbers = [0, 1, 2, 3, 10, 20, 30];
numbers.sort((a, b) => a - b);
console.log(numbers);array sort js
arr = ['width', 'score', done', 'neither' ]
arr.sort() // results to ["done", "neither", "score", "width"]
arr.sort((a,b) => a.localeCompare(b))
// if a-b (based on their unicode values) produces a negative value,
// a comes before b, the reverse if positive, and as is if zero
//When you sort an array with .sort(), it assumes that you are sorting strings
//. When sorting numbers, the default behavior will not sort them properly.
arr = [21, 7, 5.6, 102, 79]
arr.sort((a, b) => a - b) // results to [5.6, 7, 21, 79, 102]
// b - a will give you the reverse order of the sorted items
//this explnation in not minejavascript sort
homes.sort((a, b) => parseFloat(a.price) - parseFloat(b.price));how to sort an array
import java.util.Scanner;
public class JavaExample
{
public static void main(String[] args)
{
int count, temp;
//User inputs the array size
Scanner scan = new Scanner(System.in);
System.out.print("Enter number of elements you want in the array: ");
count = scan.nextInt();
int num[] = new int[count];
System.out.println("Enter array elements:");
for (int i = 0; i < count; i++)
{
num[i] = scan.nextInt();
}
scan.close();
for (int i = 0; i < count; i++)
{
for (int j = i + 1; j < count; j++) {
if (num[i] > num[j])
{
temp = num[i];
num[i] = num[j];
num[j] = temp;
}
}
}
System.out.print("Array Elements in Ascending Order: ");
for (int i = 0; i < count - 1; i++)
{
System.out.print(num[i] + ", ");
}
System.out.print(num[count - 1]);
}
}Copyright © 2021 Codeinu
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