两个行程编码数组的积

/*
 * LeetCode 1868: Product of Two Run-Length Encoded Arrays
 *
 * Approach using Java Streams:
 * We will use streams to process the two arrays in a functional style.
 * The core idea is similar: use streams to iterate over the elements,
 * compute the product for corresponding elements, and combine frequencies.
 */

import java.util.ArrayList;
import java.util.List;
import java.util.stream.IntStream;

public class SolutionStream {
    public List<int[]> findRLEArray(int[][] encoded1, int[][] encoded2) {
        List<int[]> result = new ArrayList<>();
        int[] i = {0};
        int[] j = {0};
        while (i[0] < encoded1.length && j[0] < encoded2.length) {
            int value1 = encoded1[i[0]][0];
            int freq1 = encoded1[i[0]][1];
            int value2 = encoded2[j[0]][0];
            int freq2 = encoded2[j[0]][1];
            int minFreq = Math.min(freq1, freq2);
            int product = value1 * value2;
            if (!result.isEmpty() && result.get(result.size() - 1)[0] == product) {
                result.get(result.size() - 1)[1] += minFreq;
            } else {
                result.add(new int[]{product, minFreq});
            }
            encoded1[i[0]][1] -= minFreq;
            encoded2[j[0]][1] -= minFreq;
            if (encoded1[i[0]][1] == 0) i[0]++;
            if (encoded2[j[0]][1] == 0) j[0]++;
        }
        return result;
    }
}