用java写这道题Given a 0-indexed 2D grid of size m x n, you should find the matrix answer of size m x n. The value of each cell (r, c) of the matrix answer is calculated in the following way: Let topLeft[r][c] be the number of distinct values in the top-left diagonal of the cell (r, c) in the matrix grid. Let bottomRight[r][c] be the number of distinct values in the bottom-right diagonal of the cell (r, c) in the matrix grid. Then answer[r][c] = |topLeft[r][c] - bottomRight[r][c]|. Return the matrix answer. A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end. A cell (r1, c1) belongs to the top-left diagonal of the cell (r, c), if both belong to the same diagonal and r1 < r. Similarly is defined bottom-right diagonal.

以下是Java代码实现：

class Solution {
    public int[][] diagonalDifference(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][] answer = new int[m][n];
        
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                Set<Integer> topLeft = new HashSet<>();
                Set<Integer> bottomRight = new HashSet<>();
                
                // Calculate top-left diagonal
                for (int i = r-1, j = c-1; i >= 0 && j >= 0; i--, j--) {
                    topLeft.add(grid[i][j]);
                }
                
                // Calculate bottom-right diagonal
                for (int i = r+1, j = c+1; i < m && j < n; i++, j++) {
                    bottomRight.add(grid[i][j]);
                }
                
                // Calculate answer
                answer[r][c] = Math.abs(topLeft.size() - bottomRight.size());
            }
        }
        
        return answer;
    }
}

在这个解法中，我们首先定义一个大小为m x n的answer数组，并初始化为0。然后，我们遍历每个单元格(r, c)。对于每个单元格，我们使用两个嵌套循环计算它的top-left和bottom-right对角线中不同的值。我们使用两个HashSet来存储这些值，因为HashSet可以自动去重。最后，我们计算这个单元格的答案，并将其放入answer数组中。最后，我们返回answer数组。