Swift:第二次出现 indexOf

Question

let numbers = [1,3,4,5,5,9,0,1]

要查找前 5，请使用:

numbers.indexOf(5)

如何找到第二次出现的地方？

Answer 1

• 列表项

您可以对剩余数组切片中的元素索引执行另一次搜索，如下所示:

编辑/更新:Swift 5.2 或更高版本

extension Collection where Element: Equatable {
    /// Returns the second index where the specified value appears in the collection.
    func secondIndex(of element: Element) -> Index? {
        guard let index = firstIndex(of: element) else { return nil }
        return self[self.index(after: index)...].firstIndex(of: element)
    }
}

---

extension Collection {
    /// Returns the second index in which an element of the collection satisfies the given predicate.
    func secondIndex(where predicate: (Element) throws -> Bool) rethrows -> Index? {
        guard let index = try firstIndex(where: predicate) else { return nil }
        return try self[self.index(after: index)...].firstIndex(where: predicate)
    }
}

测试:

let numbers = [1,3,4,5,5,9,0,1]
if let index = numbers.secondIndex(of: 5) {
    print(index)    // "4\n"
} else {
    print("not found")
}    
if let index = numbers.secondIndex(where: { $0.isMultiple(of: 3) }) {
    print(index)    // "5\n"
} else {
    print("not found")
}