[英]How do you find the sum of certain elements in an array in java
我正在尝试获取sumSection
来总结跑步者呼吁的元素,但是不确定如何做到这一点?
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;
}
}
这是跑步者:
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。 我已经尝试了所有我在Java中知道的内容,但是不明白如何使用sumSection
获取数组中0-3的总和
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();
}
这正好从start
到stop
(独家),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
并且应该没有问题。
for (int i = start; (i < numArray.length) && (i <= stop); i++) {
sum += numArray[i];
}
您需要另一种类型的循环,而不是:
for (int i : numArray)
更好的方法是:
int sum = 0;
if(stop <= array.length && start < stop) {
for(int i = start; i < stop; i++) {
sum += array[i];
}
}
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