Scala Spark中两个列表/数组的比较实现方案问询

我有两个列表：

# 耗时
x1 = List(10, 20, 30, 40, 50)

# 分配时长
y1 = List(15, 30)

以下是更多示例：

+-------------------+----------------------+----------+
|       time_taken  |       time_alloted   | result   |
+-------------------+----------------------+----------+
|         160-80-40 |            [20,40,80]|        80|
|            520-90 |           [45,90,130]|       130|
|    60-120-240-480 |              [30,120]|       120|
| 10-10-10-20-20-30 |               [12,30]|        12|
| 10-20-30-40-50-60 |               [12,30]|        30|
| 45-40-40-40-40-40 |           [45,90,130]|        45|
+-------------------+----------------------+----------+

我需要按照以下规则比较time_taken列和time_alloted列中的秒数：

• 若time_taken中的元素能被time_alloted中的元素（从大到小遍历）整除，则取最大的那个时长，例如结果为80（3个80单位和1个40单位，选取80）
• 若time_taken中的元素大于time_alloted中的最大元素，则将time_alloted中的元素乘以2、3、4……直到能被整除（如130*5=520）
• 若元素无法整除，则检查哪个元素的余数更小

本质上，我希望比较ListA（含1到n个元素）和ListB（含1到m个元素）中的每个元素，得到合适的结果。

补充说明 #

我目前使用SQL进行计算，现寻求基于数据结构或高阶函数的解决方案。
我已从两个列表中计算出大小、最大值、最小值，并将每个元素存储在单独的列中：

• time_taken列：tim_size, tim_max, tim_min, dt1, dt2, dt3
• time_alloted列：alt_size, alt_max_sec, alt_min_sec, t1, t2, t3

当前SQL逻辑如下：

when tim_size ==1 and alt_size == 1 then alt_max_sec 
when mod(tim_min, alt_max_sec) == 0 then alt_max_sec 
when tim_size ==1 and alt_size == 2 and MOD(tim_max, alt_max_sec) == 0 then alt_max_sec
when tim_size ==1 and alt_size == 2 and MOD(tim_max, alt_min_sec) == 0 then alt_min_sec
when tim_size ==1 and alt_size == 3 and MOD (tim_max, alt_max_sec) == 0 then alt_max_sec
when tim_size ==1 and alt_size == 3 and MOD (tim_max, t2) == 0 then t2
when tim_size ==1 and alt_size == 3 and MOD (tim_max, alt_min_sec) == 0 then alt_min_sec
when tim_size ==2 and alt_size == 1 then alt_max_sec
when tim_size ==2 and alt_size == 2 and MOD(tim_max, alt_max_sec) == 0 and MOD(tim_min, alt_max_sec) == 0 then alt_max_sec
when tim_size ==2 and alt_size == 2 and MOD(tim_max, alt_min_sec) == 0 and MOD(tim_min, alt_min_sec) == 0 then alt_min_sec
when tim_size ==2 and alt_size == 2 and MOD(tim_min, alt_min_sec) == 0 then alt_min_sec
when tim_size ==2 and alt_size == 3 and MOD(tim_max, alt_max_sec) == 0 and MOD(tim_min, alt_max_sec) == 0 then alt_max_sec
when tim_size ==2 and alt_size == 3 and MOD(tim_max, t2) == 0 and MOD(tim_min, t2) == 0 then t2
when tim_size ==2 and alt_size == 3 and MOD(tim_max, alt_max_sec) == 0 and MOD(tim_min, t2) == 0 and ((tim_max/alt_max_sec) > (tim_min/t2)) THEN alt_max_sec
WHEN tim_size ==2 and alt_size == 3 and (tim_max/t2) == 1 and (tim_min/t1) == 1 then alt_min_sec
WHEN tim_size ==3 and alt_size == 3 and mod(tim_max, alt_max_sec) == 0 and mod(dt2, alt_max_sec) == 0 then alt_max_sec
WHEN tim_size ==3 and alt_size == 3 and mod(tim_max, alt_max_sec) == 0 and mod(dt2, alt_max_sec) == 0 then alt_max_sec
WHEN tim_size ==3 and alt_size == 2 and MOD(tim_max, alt_max_sec) == 0 and mod(dt2, alt_max_sec) == 0 then alt_max_sec
when alt_size == 3 and mod(tim_min, t2) == 0 then t2
else 'TBC' end as result

内容的提问来源于stack exchange，提问作者nagraj036