Here's one way to divide an array list into equivalent segments when the size of the array is unknown:
Determine how many segments you want to divide the array into. This could be specified by the user or calculated based on some other parameter.
Use the ArrayList.size() method to get the current size of the array.
Divide the size of the array by the number of segments you want to create. If the result is not a whole number, round up to the nearest integer.
Use a loop to iterate over the array and create sub-lists of equal size. In each iteration, take a slice of the original array that is equal to the rounded-up segment size, using the subList() method.
Add each sub-list to a new ArrayList or other data structure that can hold a collection of lists.
If there are any remaining elements in the original array after all the segments have been created, add them to the last sub-list.
Here's some sample code to demonstrate this algorithm:
import java.util.ArrayList;
public class ArraySegmenter {
public static void main(String[] args) {
ArrayList<Integer> originalArray = new ArrayList<>();
// fill originalArray with some data
int numSegments = 4;
int originalSize = originalArray.size();
int segmentSize = (int) Math.ceil((double) originalSize / numSegments);
ArrayList<ArrayList<Integer>> segmentedArray = new ArrayList<>();
int startIndex = 0;
int endIndex = Math.min(startIndex + segmentSize, originalSize);
while (endIndex <= originalSize) {
ArrayList<Integer> segment = new ArrayList<>(originalArray.subList(startIndex, endIndex));
segmentedArray.add(segment);
startIndex = endIndex;
endIndex = Math.min(startIndex + segmentSize, originalSize);
}
// add any remaining elements to the last segment
if (startIndex < originalSize) {
ArrayList<Integer> lastSegment = new ArrayList<>(originalArray.subList(startIndex, originalSize));
segmentedArray.add(lastSegment);
}
// now segmentedArray contains numSegments sub-lists of equal size (except possibly the last one)
}
}
Asked: 2023-07-15 11:05:34 +0000
Seen: 14 times
Last updated: Jul 15 '23