list - 如何在函数中返回带有列表参数的 MutableMap(HashMap) 以查找列表中每个元素的频率（Kotlin）

Question

fun main() {
    val list = listOf("B", "A", "A", "C", "B", "A")
    print(findfrequency(list))
}

fun <T> findfrequency(list: T): MutableMap<String, Int> {
    val frequencyMap: MutableMap<String, Int> = HashMap()
 
    for (s in list) {
        var count = frequencyMap[s]
        if (count == null) count = 0
        frequencyMap[s] = count + 1
    }
    
    return frequencyMap
}

Answer 1

解决方法：不需要对变量列表进行泛型类型声明，只需添加<String>函数参数即可。这是最终的程序：

fun main() {
    val list_String = listOf("B", "A", "A", "C", "B", "A")
    println(findfreq(list_String))
}

fun findfreq(list: List<String>): MutableMap<String, Int> {
    val frequencyMap: MutableMap<String, Int> = HashMap()
    
    for(i in list) {
        var count = frequencyMap[i]
        if(count == null) count = 0
        frequencyMap[i] = count + 1
    }
    
    return frequencyMap
}

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Answer 2

当你将一个字符串列表放入一个类型为 T 的函数中时，它的类型在运行时是未知的，你正试图迭代这个未知的类型。因此，您必须指明该对象可用于与之交互。

fun main() {
    val list = listOf("B", "A", "A", "C", "B", "A")
    print(findfrequency(list))
}

fun <T : Iterable<String>> findfrequency(list: T): MutableMap<String, Int> {
    val frequencyMap: MutableMap<String, Int> = HashMap()

    for (s in list) {
        var count = frequencyMap[s]
        if (count == null) count = 0
        frequencyMap[s] = count + 1
    }

    return frequencyMap
}

在下面的示例中，我们不再像上面那样依赖 String 类型。但是，在输出中，我们会得到一个这种类型的 Map。

fun <T : Iterable<S>, S> findfrequency(list: T): MutableMap<S, Int> {
    val frequencyMap: MutableMap<S, Int> = HashMap()

    for (s in list) {
        var count = frequencyMap[s]
        if (count == null) count = 0
        frequencyMap[s] = count + 1
    }

    return frequencyMap
}