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Radioactive
Asked: 2024-07-02 02:13:37 +0800 CST

如何返回 Array<Array<String>> 的后代的相同元素？

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我有一个字符串数组，我需要返回后代的相同元素（不是索引方式，而是自身方式）。

let arrayArrayStrings: [[String]] = [
["a", "b", "e", "q"],
["a", "c", "q", "e"],
["a", "q", "d", "e"]]

我希望它返回["a", "e", "q"]。我试图做到这一点时大脑一片混乱。我怎样才能有效地做到这一点？

arrays

1 个回答

  1. Best Answer

    Set具有强大的方法来确定两个集合的关系，因此一个可能的解决方案是将内部数组映射到Set并调用intersection。

    let arrayArrayStrings: [[String]] = [
["a", "b", "e"],
["a", "c", "e"],
["a", "d", "e"]]

let arraySetStrings = arrayArrayStrings.map(Set.init)
var commonElements = arraySetStrings[0]
for i in 1..<arraySetStrings.count {
    commonElements.formIntersection(arraySetStrings[i])
}
print(commonElements.sorted())

    检查元素位置的另一种方法是使用difference(fromAPIArray

    let arrayArrayStrings: [[String]] = [
["a", "b", "e"],
["a", "c", "e"],
["a", "d", "e"]]

var commonElements = arrayArrayStrings[0]
var indicesToRemove = IndexSet()
for i in 1..<arrayArrayStrings.count {
    let diff = commonElements.difference(from: arrayArrayStrings[i])
    diff.removals.forEach { removal in
        if case .remove(offset: let offset, element: let element, associatedWith: let associatedWith) = removal {
            indicesToRemove.insert(offset)
        }
    }
}
commonElements.remove(atOffsets: indicesToRemove)
print(commonElements)
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