予有字符串两个阵列，相同长度的不是必须的，我想找到的所有可能的套的来自阵列的两个值之间的组合，而不偏离任一阵列重复。 例如，给定的阵列： {A1，A2，A3} {B1，B2} 我要的结果如下设置： {（A1，B1），（A2，B2）} {（A1，B1），（A3，B 2）} {（A1，B2），（A2，B1）} {（A1，B2），（A3，B 1）} {（A2，B1），（A3，B 2）} {（A2，B2），（A3，B 1）}

I have two arrays of strings, not necessarily of the same length, I want to find all the possible "sets" of combinations between two values from the arrays, without repeats from either array. For example, given the arrays: { "A1", "A2", "A3" } { "B1", "B2" } The result I want is the following sets: { ("A1", "B1"), ("A2", "B2") } { ("A1", "B1"), ("A3", "B2") } { ("A1", "B2"), ("A2", "B1") } { ("A1", "B2"), ("A3", "B1") } { ("A2", "B1"), ("A3", "B2") } { ("A2", "B2"), ("A3", "B1") }

我的总的方向是建立递归函数，它作为参数的两个数组并删除每一个选择字符串的时间，自称直到数组是空的，但是我有点担心性能问题（我需要运行在大约千双字符串数组）本code。 任何人都可以直接我对一个有效的方法来做到这一点？

My general direction is to create recursive function that takes as a parameter the two arrays and removes each "chosen" strings at a time, calling itself until either array is empty, however I'm kinda worried about performance issues (I need to run this code on about a 1000 pairs of string arrays). Can anyone direct my towards an efficient method to do this?

推荐答案

这可能是有益的，觉得两个数组作为表的两面：

It might be beneficial to think of the two arrays as sides of a table:

        A1      A2      A3
---+-------+-------+-------+
B1 | B1,A1 | B1,A2 | B1,A3 |
---+-------+-------+-------+
B2 | B2,A1 | B2,A2 | B2,A3 |
---+-------+-------+-------+

此意味着嵌套在另一个循环，一个循环为行，另一个用于列。这会给你的初始设置对：

This implies a loop nested within another, one loop for the rows and the other for the columns. This will give you the initial set of pairs:

{B1,A1} {B1,A2} {B1,A3} {B2,A1} {B2,A2} {B2,A3}

然后，它是建立该初始集合的组合的问题。可以想像的组合同样地，与该组对的两个行和列：

Then it is a matter of building up the combinations of that initial set. You can visualise the combinations similarly, with the set of pairs for both the rows and columns:

      B1,A1 B1,A2 B1,A3 B2,A1 B2,A2 B2,A3
-----+-----+-----+-----+-----+-----+-----+
B1,A1|     |  X  |  X  |  X  |  X  |  X  |
-----+-----+-----+-----+-----+-----+-----+
B1,A2|     |     |  X  |  X  |  X  |  X  |
-----+-----+-----+-----+-----+-----+-----+
B1,A3|     |     |     |  X  |  X  |  X  |
-----+-----+-----+-----+-----+-----+-----+
B2,A1|     |     |     |     |  X  |  X  |
-----+-----+-----+-----+-----+-----+-----+
B2,A2|     |     |     |     |     |  X  |
-----+-----+-----+-----+-----+-----+-----+
B2,A3|     |     |     |     |     |     |
-----+-----+-----+-----+-----+-----+-----+

同样可以完成此一对嵌套循环（提示：你的内循环的范围将由外部循环的值来确定）

Again this can be accomplished with a pair of nested loops (hint: your inner loop's range will be determined by the outer loop's value).