java - 你如何在java中找到数组中某些元素的总和

Question

我正在尝试让 sumSection 总结运行者调用的元素，但不确定如何做？

package com.company;

public class processser
{
    //instance variables and constructors could be present, but are not necessary

    //sumSection will return the sum of the numbers
    //from start to stop, not including stop
    public static int sumSection(int[] numArray, int start, int stop)
    {
        int sum = 0;
        {
            for (int i : numArray)
                sum += i;
        }
        return sum ;
    }

    //countVal will return a count of how many times val is present in numArray
    public static int countVal(int[] numArray, int val)
    {
        int count = 0;
        for (int item : numArray)
        {
            if (item == val)
            {
                count = count + 1;
            }
        }
        return count;
    }
}

这是运行者:

package com.company;

import static java.lang.System.*;
import java.lang.Math;
import java.util.Arrays;

public class Main
{
    public static void main(String args[])
    {
        int[] theRay = {2,4,6,8,10,12,8,16,8,20,8,4,6,2,2};

        out.println("Original array : "+ Arrays.toString(theRay));

        out.println("Sum of 0-3: " + processser.sumSection(theRay, 0, 3));

    }
}

我正在尝试获取数组 0-3 中位置的总和。我已经尝试了我在 Java 中所知道的一切，但不明白如何使用 sumSection

获取数组中 0-3 的总和

Answer 1

您可以使用 Java 8 Streams:

static int sumSection(int[] numArray, int start, int stop) {
    return IntStream.range(start, stop).map(i -> numArray[i]).sum();
}

这来自 start至 stop (独占)，所以如果你有:

int[] theRay = {2,4,6,8,10,12,8,16,8,20,8,4,6,2,2};
sumSection(theRay, 0, 3);

它会像这样工作:

IntStream.range(0, 3) -> [0, 1, 2]
[0, 1, 2].map(i -> numArray[i]) -> [2, 4, 6]
[2, 4, 6].sum() -> 12

只要确保start < stop和 stop <= numArray.length你应该没有问题。