我有一个填充了整数的arrayList,我需要遍历这个arrayList,添加数字直到达到一个阈值,然后将得到的总和放到第二个arrayList槽中,然后移回到我在原始的arrayList,继续迭代和求和,直到达到阈值,然后将其放在第二个槽中,依此类推,直到所有40个原始项已经求和并放入一个较小的arrayList中。
我想过使用两个嵌套循环,但我无法让两个循环一起工作。
我是Java的新手,并且不知道如何做到这一点。任何建议都会有所帮助。
这是我到目前为止所做的:
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
答案 0 :(得分:2)
实际上只需一个循环就可以做到这一点。你不需要搬回去,只要呆在原地并继续前进。
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;
}
您可以毫无困难地将输入参数从ArrayList更改为int []或Integer []。万岁为每个循环!
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());
}
答案 1 :(得分:0)
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));
}
}
}
<强>输出强>
15
13
17
21
12
13
14
答案 2 :(得分:0)
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]