val list = listOf("A", "A", "B", "A", "A", "A", "A", "C", "C", "A")

val result = list.foldIndexed(mutableListOf<MutableList<String>>()) { index, acc, value ->
  when {
    index > 0 && list[index - 1] == value -> acc.last().add(value)
    else                                  -> acc.add(mutableListOf(value))
  }
  acc
}

result.forEach(::println)

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Kotlin Playground

这里有一个使用序列的解决方案

fun collectSame(src: List<String>): Sequence<Pair<String, List<String>>> = sequence {
    var currentGroupKey: String? = null
    val collector = mutableListOf<String>()
    src.forEach {
        if(currentGroupKey==null || currentGroupKey != it) {
            currentGroupKey = it
            if(collector.isNotEmpty()) {
                yield(currentGroupKey!! to collector)
                collector.clear()
            }
        }
        collector.add(it)
    }
}

fun main(args: Array<String>) {
    val answer = collectSame(listOf("A","A","B","A","A","A","A","C","C","A"))
    answer.forEach {
        println(it.second)
    }
}

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输出

[A, A]
[B]
[A, A, A, A]
[C, C]

型
可能有一些方法比我的grouping解决方案更好地做到这一点。