遍历数组，求和并将结果放入新数组答案

【问题标题】：Iterate through array, sum and put results in new array遍历数组，求和并将结果放入新数组
【发布时间】：2014-09-23 03:56:23
【问题描述】：

我有一个用整数填充的 arrayList，我需要遍历这个 arrayList，将数字相加直到达到阈值，然后将结果总和放入第二个 arrayList 槽，然后回到我在原始arrayList，不断迭代和求和，直到达到阈值，然后将其放入第二个槽中，依此类推，直到将所有40个原始项求和并放入较小的arrayList。

我曾想过使用两个嵌套循环，但我无法让这两个循环一起工作。

我对 Java 很陌生，不知道该怎么做。任何建议都会有所帮助。

这是我目前所拥有的：

int threshold = 12;
int sumNum = 0;
int j = 0;
int arr1[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};
for (int i = 0; i < arr1.length; i++) {
  while (sumNum <= threshold) {
  sumNum += arr[j];
  j++
}//end while

}//end for

【问题讨论】：

标签： java arrays arraylist iteration

【解决方案1】：

实际上，您只需一个循环即可完成此操作。您无需后退，只需原地不动，继续求和。

public static ArrayList<Integer> sums(ArrayList<Integer> arr, int threshold){
    ArrayList<Integer> sumArr = new ArrayList<Integer>();
    int s = 0;  //Sum thus far, for the current sum
    for(int i : arr){
        s += i;             //Add this element to the current sum
        if(s >= threshold){ //If the current sum has reached/exceeded the threshold
            sumArr.add(s);  //Add it to the sumArray, reset the sum to 0.
            s = 0;
        }
    }
    return sumArr;
}

您可以毫无问题地将输入参数从 ArrayList 更改为 int[] 或 Integer[]。 for-each 循环万岁！

使用上述代码：

public static void main(String[] args){
    ArrayList<Integer> i = new ArrayList<Integer>();
    //create arraylist 1..20
    for(int x = 1; x <= 20; x++){
        i.add(x);
    }

    System.out.println(sums(i).toString());
}

【讨论】：

我会保证这个答案。
注意，这里假定arr 是一个实际的ArrayList，而不是问题中提供的原始数组。
没错，但引用第一句话：“我有一个用整数填充的arrayList...”
鉴于问题的性质，我想说可能会有一些混乱。
似乎很可能。那（array vs ArrayList）是一个单独的问题，在这里和其他地方都有很好的记录。同时，代码可以使用任何一种方式，只需更改第一个参数的类型以适合您的喜好。

【解决方案2】：

package com.test;

import java.util.ArrayList;
import java.util.List;

public class Test {
    public static void main(String[] args) {
        int threshold = 12;
        int sumNum = 0;
        int j = 0;
        int arr1[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 };
        List myList = new ArrayList();

        for (int i = 0 ; i < arr1.length ; i++) {
            sumNum += i;
            if (sumNum >= threshold) {
                myList.add(sumNum);
                sumNum = 0;
            }
        }

        for (int a = 0 ; a < myList.size() ; a++) {
            System.out.println(myList.get(a));
        }
    }
}

输出

【讨论】：

【解决方案3】：

这样的事情怎么样：

/**
 * Sequentially adds numbers found in the source array until the sum >= threshold.
 * <p/>
 * Stores each threshold sum in a separate array to be returned to the caller
 * <p/>
 * Note that if the last sequence of numbers to be summed does not meet the threshold,
 * no threshold sum will be added to the result array
 * 
 * @param numbersToSum The source number list
 * @param threshold The threshold value that determines when to move on to 
 * the next sequence of numbers to sum
 * 
 * @return An array of the calculated threshold sums
 */

public static Integer[] sumWithThreshold(int[] numbersToSum, int threshold)
{
    List<Integer> thresholdSums = new ArrayList<Integer>();

    if (numbersToSum != null)
    {
        int workingSum = 0;

        for (int number: numbersToSum)
        {
            workingSum = workingSum + number;

            if (workingSum >= threshold)
            {
                thresholdSums.add(workingSum);
                workingSum = 0;
            }
        }
    }

    return thresholdSums.toArray(new Integer[thresholdSums.size()]);
}

public static void main(String[] args)
{
    int[] testNumbers = 
    {
            1,2,3,4,5,6,7,8,9,10,
            11,12,13,14,15,16,17,18,19,20,
            21,22,23,24,25,26,27,28,29,30,
            31,32,33,34,35,36,37,38,39,40};

    int[] thresholds = {1, 42, 100, 200};

    for (int threshold: thresholds)
    {
        System.out.println("Threshold sums for threshold = " + threshold + ":\n" + Arrays.toString(sumWithThreshold(testNumbers, threshold)));
    }
}

产生以下输出：

Threshold sums for threshold = 1:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40]
Threshold sums for threshold = 42:
[45, 46, 45, 54, 63, 47, 51, 55, 59, 63, 67, 71, 75, 79]
Threshold sums for threshold = 100:
[105, 105, 115, 110, 126, 105, 114]
Threshold sums for threshold = 200:
[210, 225, 231]

【讨论】：