Kotlin递归函数比较两个数组并保留原始数组的内容

Question

我需要使用递归函数比较两个 MutableList，requiredDevices 和 providedDevices。这两个 MutableList 是一个 MutableList 的元素，originalArray 也是如此。如果有匹配，则将requiredDevices和providedDevices中匹配的元素一个一个取出，剩下的分别添加到list1和list2中。我怎样才能做到这一点并保留 requiredDevices 和 providedDevices 和 originalArray 的内容。例如：

originalArray=[requiredDevices, providedDevices]
           =[["A","B","C"],["B,"C","A","A","D"]]

requiredDevices=["A","B","C"]
providedDevices=["B,"C","A","A","D"]

，递归后：

list1=[] (empty array)
list2=["A","D"]

originalArray=[[],["A","D"]]

尝试编写递归函数并将 requiredDevices 和 providedDevices 作为参数。递归函数返回一个包含 list1 和 list2 的列表 temp。但是 requiredDevices 和 providedDevices 也被修改了。

val temp = mutableListOf<MutableList<String>>()
fun compareArray(requiredDevices: MutableList<String>, providedDevices: MutableList<String>): List<MutableList<String>> {

    for (i in 1 until requiredDevices.size) {
        for (j in 0 until providedDevices.size) {

            try {
                if (requiredDevices[i] == providedDevices[j]) {
                    requiredDevices.removeAt(i)
                    providedDevices.removeAt(j)

                    compareArray(requiredDevices, providedDevices)

                    temp.add(requiredDevices)
                    temp.add(providedDevices)
                }
            } catch (e: IndexOutOfBoundsException) {
            }
        }
    }

    return temp.distinct()
}

我想保留 originalArray 的内容，并且仍然得到 list1 和 list2 相同的结果。如下：

originalArray=[["A","B","C"],["B,"C","A","D"]]

Answer 1

我并没有真正遵循您的算法试图做什么。根据您的评论，我认为这可以解决问题。这会将两个起始列表复制到新列表中，然后从副本中逐一删除项目，因此如果有多余的项目，它们将保留它们。

fun <T> removeCommonValues(listA: List<T>, listB: List<T>): Pair<List<T>, List<T>> {
    val outA = listA.toMutableList()
    val outB = listB.toMutableList()
    for (i in (outA.size - 1) downTo 0){
        for (j in (outB.size - 1) downTo 0){
            if (outA[i] == outB[j]){
                outA.removeAt(i)
                outB.removeAt(j)
                break
            }
        }
    }

    return outA to outB
}

测试：

fun main() {
    val required = listOf("A", "B", "C", "C")
    val available = listOf("B", "C", "A", "D", "D")
    println(removeCommonValues(required, available))
}

// Outputs ([C], [D, D])