Question

我有一系列正/负的整数

int[] numbers = new int[10];
numbers[0] = 100;
numbers[1] = -34200;
numbers[2] = 3040;
numbers[3] = 400433;
numbers[4] = 500;
numbers[5] = -100;
numbers[6] = -200;
numbers[7] = 532;
numbers[8] = 6584;
numbers[9] = -945;

现在，我想针对这个数组测试另一个int，并返回最接近int的数字。

例如，如果我使用数字490，我会从数字500返回第4项，这样做的最佳方法是什么？

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

这不起作用。有关一个好方法的任何建议吗？

Answer 1

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];

始终使用您正在考虑的第一个元素初始化您的最小/最大函数。使用诸如Integer.MAX_VALUE或Integer.MIN_VALUE之类的东西是获得答案的天真方式;如果您稍后更改数据类型（哎呀，MAX_LONG和MAX_INT非常不同！）或者您将来想要编写通用min/max方法，它就不会很好用对于任何数据类型。

Answer 2

在 Java 8 ：

中

List<Integer> list = Arrays.stream(numbers).boxed().collect(Collectors.toList());

int n = 490;

int c = list.stream()
            .min(Comparator.comparingInt(i -> Math.abs(i - n)))
            .orElseThrow(() -> new NoSuchElementException("No value present"));

最初，您可以使用List代替Array（列表具有更多功能）。

Answer 3

你非常接近。我认为'distance'的初始值应该是一个大数而不是0.并使用cdistance的绝对值。

Answer 4

cdistance = numbers[c] - myNumber。你没有采取差异的绝对值。如果myNumber远大于numbers[c]或numbers[c]为负数，则比较将注册为“最小差异”。

以numbers[c] = -34200为例。 numbers[c] - myNumber将为-34690，远低于distance。

此外，您应该将distance初始化为较大的值，因为在开始时未找到任何解决方案。

Answer 5

一个语句块初始化并设置最接近的匹配。此外，如果找不到最接近的匹配，则返回-1（空数组）。

 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;
}

Answer 6

int valueToFind = 490;

Map<Integer, Integer> map = new HashMap();

for (int i = 0, i < numbers.length; i++){
    map.put(Math.abs(numbers[i] - valueToFind), numbers[i]);
}

List<Integer> keys = new ArrayList(map.keySet());
Collections.sort(keys);

return map.get(keys.get(0));

Answer 7

我这样做是为了我的课程作业，我把它编程为Ready to Program Java，很抱歉，如果它有点令人困惑。

// 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 8

public int getClosestToTarget(int target, int[] values) {

    if (values.length < 1)
        throw new IllegalArgumentException("The values should be at least one element");
    if (values.length == 1) {
        return values[0];
    }
    int closestValue = values[0];
    int leastDistance = Math.abs(values[0] - target);
    for (int i = 0; i < values.length; i++) {
        int currentDistance = Math.abs(values[i] - target);
        if (currentDistance < leastDistance) {
            closestValue = values[i];
            leastDistance = currentDistance;
        }
    }
    return closestValue;
}

Answer 9

您可以调整旧的二进制搜索并有效地实现它。

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 10

科特琳很有帮助

fun List<Int>.closestValue(value: Int) = minBy { abs(value - it) }

val values = listOf(1, 8, 4, -6)

println(values.closestValue(-7)) // -6
println(values.closestValue(2)) // 1
println(values.closestValue(7)) // 8

列表无需进行顺便说一句

Answer 11

<块引用>

Kotlin - TreeSet lower 方法返回此中最大的元素设置严格小于给定元素，如果没有这样的元素，则为 null 元素。

import java.util.*

fun printLowest(number: Int) {
    val numbers = listOf(100, 90, 50, -100, -200, 532, 6584, -945)
    val lower = TreeSet(numbers).lower(number)
    println(lower)
}

printLowest(100) // Prints 90

Answer 12

public int nearnum(int myNumber,int[] numbers)
    {
        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;
            }
        }
        return numbers[idx];
    }

Answer 13

public class Main    
{
    public static void main(String[] args)
    {   
        int[] numbers = {6,5,10,1,3,4,2,14,11,12};

        for(int i =0; i<numbers.length; i++)
        {
            sum(numbers, i, numbers[i], 12, String.valueOf(numbers[i]));
        }
    }

    static void sum(int[] arr, int i, int sum, int target, String s)
    {

        int flag = 0;

        for(int j = i+1; j<arr.length; j++)
        {

            if(arr[i] == target && flag==0)
            {
                System.out.println(String.valueOf(arr[i]));
                flag =1;

            }
            else if(sum+arr[j] == target)
            { 
                System.out.println(s+" "+String.valueOf(arr[j]));

            }
            else
            {
                sum(arr, j, sum+arr[j], target, s+" "+String.valueOf(arr[j]));
            }
        }        
    }
}

Answer 14

这是我做过的事情......

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;
}

}

获得最接近数组中的数字的值

14 个答案: