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private int m, n;

    private int dfs(int[][] matrix, int[][] dp, int i, int j) {
        // 记忆化：已经计算过该位置的最长路径，直接返回
        if (dp[i][j] != 0) {
            return dp[i][j];
        }
        
        dp[i][j]++;

        
        if (i - 1 >= 0 && matrix[i - 1][j] > matrix[i][j]) {
            dp[i][j] = Math.max(dp[i][j], dfs(matrix, dp, i - 1, j) + 1);
        }
        
        if (i + 1 < m && matrix[i + 1][j] > matrix[i][j]) {
            dp[i][j] = Math.max(dp[i][j], dfs(matrix, dp, i + 1, j) + 1);
        }
        
        if (j - 1 >= 0 && matrix[i][j - 1] > matrix[i][j]) {
            dp[i][j] = Math.max(dp[i][j], dfs(matrix, dp, i, j - 1) + 1);
        }
        
        if (j + 1 < n && matrix[i][j + 1] > matrix[i][j]) {
            dp[i][j] = Math.max(dp[i][j], dfs(matrix, dp, i, j + 1) + 1);
        }
        
        return dp[i][j];
    }

    public int solve(int[][] matrix) {
        
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }

        int ans = 0;
        m = matrix.length;
        n = matrix[0].length;
        
        int[][] dp = new int[m][n];

        // 遍历矩阵的每一个单元格，作为递增路径的起点
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                ans = Math.max(ans, dfs(matrix, dp, i, j));
            }
        }
        return ans;
    }