矩阵中最大的三个菱形和

/*
 * Problem: Find the three largest unique rhombus sums in a given 2D grid using Java Streams.
 * A rhombus is defined as a square rotated 45 degrees. The sum of a rhombus is the sum of all the elements on its boundary.
 * Approach:
 * 1. Use streams to iterate through each possible center of the rhombus and calculate all possible rhombus sums.
 * 2. Store unique sums in a set and sort them in descending order.
 * 3. Return the top three unique sums.
 */
import java.util.*;
import java.util.stream.*;

public class RhombusSumsStream {
    public List<Integer> getBiggestThree(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        Set<Integer> uniqueSums = IntStream.range(0, m)
            .boxed()
            .flatMap(i -> IntStream.range(0, n)
                .boxed()
                .flatMap(j -> IntStream.range(0, Math.min(Math.min(i, j), Math.min(m - i - 1, n - j - 1)) + 1)
                    .map(size -> calculateRhombusSum(grid, i, j, size))
                    .boxed()))
            .collect(Collectors.toSet());

        return uniqueSums.stream()
            .sorted(Comparator.reverseOrder())
            .limit(3)
            .collect(Collectors.toList());
    }

    private int calculateRhombusSum(int[][] grid, int i, int j, int size) {
        int sum = 0;
        for (int k = 0; k <= size; k++) {
            sum += grid[i - k][j + k];
            sum += grid[i + k][j + k];
            sum += grid[i + k][j - k];
            sum += grid[i - k][j - k];
        }
        if (size > 0) {
            sum -= grid[i][j - size];
            sum -= grid[i][j + size];
            sum -= grid[i - size][j];
            sum -= grid[i + size][j];
        }
        return sum;
    }
}