maybe zip and then flatmap?

# announcements

[clubfan]

maybe zip and then flatmap?

[Daniel]

Thank you and let's see about this with a short unit test

[clubfan]

listOf("a", "b", "c").zip(listOf("d", "e")).flatMap(it.toList()) maybe? Haven't tested it though

[Daniel]

Very close! This gets me to <[a, d, b, e]>

[Daniel]

On another note: the second list is always -1 the size of the first one, so this would be expect

[Daniel]

list1.zip(list2).flatMap { it.toList() } + list1.takeLast(1) would do it

[clubfan]

fun <T> List<T>.merge(other : List<T>) : List<T> {
    val result = ArrayList<T>()
    for ( i in 0 .. maxOf(this.size, other.size)) {
        this.getOrNull(i)?.let { result.add(it) }
        other.getOrNull(i)?.let { result.add(it) }
    }
    return result
}

works, but it's not really idiomatic I think

[clubfan]

and you can call it like this: listOf("a", "b", "c").merge(listOf("d", "e"))

[clubfan]

or this:

fun <T> List<T>.merge(other : List<T>) : List<T> {
    return this.mapIndexed{i, el -> other.getOrNull(i)?.let { listOf(el, it) } ?: listOf(el)}.flatten() + other.drop(this.size)

}

[clubfan]

I'm pretty sure there's a better solution

[clubfan]

Scala's zipAll would be great for this

[ilya.gorbunov]

Please add your use case to this issue: https://youtrack.jetbrains.com/issue/KT-13017

[ilya.gorbunov]

Note that if the elements in lists are nullable, you won't be able to distinguish what the last null in zipAll means: either null last element or missing last element.

[Daniel]

Thank you both for your replies! Currently I am settling for the .zip sulution with the awesome require(other.size == size - 1) function beforehand!