get closest value to a number in array

Question

I have an array of positive/negative ints

int[] numbers = new int[10];
numbers[0] = 100;
numbers[1] = -34200;
numbers[2] = 3040;
numbers[3] = 400433;
numbers         


        

Answer 1

cdistance = numbers[c] - myNumber. You're not taking the absolute value of the difference. If myNumber is a lot greater than numbers[c] or if numbers[c] is negative, the comparison will register as the "minimum difference".

Take for example the case where numbers[c] = -34200. numbers[c] - myNumber would then be -34690, a lot less than the distance.

Also, you should initialize distance to a large value, as no solution has been found at the start.

Answer 2

Here is something that i did...

import javax.swing.JOptionPane;

public class NearestNumber {

public static void main(String[] arg)
{
    int[] array={100,-3420,3040,400433,500,-100,-200,532,6584,-945};

    String myNumberString =JOptionPane.showInputDialog(null,"Enter the number to test:");
    int myNumber = Integer.parseInt(myNumberString);

    int nearestNumber = findNearestNumber(array,myNumber);

    JOptionPane.showMessageDialog(null,"The nearest number is "+nearestNumber);
}

public static int findNearestNumber(int[] array,int myNumber)
{

    int min=0,max=0,nearestNumber;

    for(int i=0;i<array.length;i++)
    {
        if(array[i]<myNumber)
        {
            if(min==0)
            {
                min=array[i];
            }
            else if(array[i]>min)
            {
                min=array[i];
            }
        }
        else if(array[i]>myNumber)
        {
            if(max==0)
            {
                max=array[i];
            }
            else if(array[i]<max)
            {
                max=array[i];
            }
        }
        else
        {
            return array[i];
        }
    }

    if(Math.abs(myNumber-min)<Math.abs(myNumber-max))
    {
        nearestNumber=min;
    }
    else
    {
        nearestNumber=max;
    }

    return nearestNumber;
}

}

Answer 3

You can tweak the good old binary search and implement this efficiently.

Arrays.sort(numbers);
nearestNumber = nearestNumberBinarySearch(numbers, 0, numbers.length - 1, myNumber);

private static int nearestNumberBinarySearch(int[] numbers, int start, int end, int myNumber) {
    int mid = (start + end) / 2;
    if (numbers[mid] == myNumber)
        return numbers[mid];
    if (start == end - 1)
        if (Math.abs(numbers[end] - myNumber) >= Math.abs(numbers[start] - myNumber))
            return numbers[start];
        else
            return numbers[end];
     if(numbers[mid]> myNumber)
        return nearestNumberBinarySearch(numbers, start,mid, myNumber);
     else
         return nearestNumberBinarySearch(numbers,mid, end, myNumber);

}

Answer 4

int myNumber = 490;
int distance = Math.abs(numbers[0] - myNumber);
int idx = 0;
for(int c = 1; c < numbers.length; c++){
    int cdistance = Math.abs(numbers[c] - myNumber);
    if(cdistance < distance){
        idx = c;
        distance = cdistance;
    }
}
int theNumber = numbers[idx];

Always initialize your min/max functions with the first element you're considering. Using things like Integer.MAX_VALUE or Integer.MIN_VALUE is a naive way of getting your answer; it doesn't hold up well if you change datatypes later (whoops, MAX_LONG and MAX_INT are very different!) or if you, in the future, want to write a generic min/max method for any datatype.

Answer 5

I did this as an assignment for my course, and I programmed it in Ready to Program Java, so sorry if its a bit confusing.

// The "Ass_1_B_3" class.
import java.awt.*;
import hsa.Console;

public class Ass_1_B_3
{
    static Console c;           // The output console

    public static void main (String[] args)
    {
        c = new Console ();

        int [] data = {3, 1, 5, 7, 4, 12, -3, 8, -2};
        int nearZero = 0;
        int temp = 0;
        int temp2 = data[0];

        for (int i = 0; i < data.length; i++)
        {
            temp = Math.abs (data[i]);
            nearZero = temp2;   
            if (temp < temp2)
            {
                temp2 = temp;
                nearZero = data[i];
            }


        }

        c.println ("The number closest to zero is: " + nearZero);

        // Place your program here.  'c' is the output console
    } // main method
} // Ass_1_B_3 class

Answer 6

One statement block to initialize and set the closest match. Also, return -1 if no closest match is found (empty array).

 protected int getClosestIndex(final int[] values, int value) {
    class Closest {
        Integer dif;
        int index = -1;
    };
    Closest closest = new Closest();
    for (int i = 0; i < values.length; ++i) {
        final int dif = Math.abs(value - values[i]);
        if (closest.dif == null || dif < closest.dif) {
            closest.index = i;
            closest.dif = dif;
        }
    }
    return closest.index;
}