Question

我的操场上有以下代码：

let array = [3,3]
let first = array.first!
let last = array.last!

let indices = [array.index(of: first)!, array.index(of: last)!]
print(indices) // This prints [0,0]

据我所知，“index（of :)”方法只是从数组中获取第一个匹配实例，效率更高，但我想知道是否有办法根据事实获取最后一个索引我从“array.last”获得了值。

此外，如果我有以下内容：

let lotsOfThrees = [3,3,3,3,3,3,3]
let fourthThree = lotsOfThrees[3]
// Write code to return the index of "fourthThree" (possibly based on memory address)

我想知道是否有办法根据内存地址执行此操作，但老实说不确定。

Answer 1

您可以通过反转数组，然后获取第一个出现的索引来获取元素的 last 索引。然后原始数组中的最后一个元素索引（反向数组中第一次出现的索引） - （数组的大小） - 1.将它放在扩展方法中以增加乐趣。

extension Array<T> {   
    func lastIndex(of item: T) -> Int? {
        if let lastIndex = self.reverse().index(of: item) {
            return self.count() - lastIndex - 1
        } else {
            return nil
        }
    }
}

Answer 2

我建议使用enumerated()和filter将索引与您要查找的值配对：

let lotsOfThrees = [3, 3, 3, 3, 3, 3, 3]
let threesAndIndices = lotsOfThrees.enumerated().filter { $1 == 3 }
print(threesAndIndices)

[(offset: 0, element: 3), (offset: 1, element: 3), (offset: 2, element: 3), (offset: 3, element: 3), (offset: 4, element: 3), (offset: 5, element: 3), (offset: 6, element: 3)]

// find index of last three
print(threesAndIndices.last!.offset)

// find index of 4th three
print(threesAndIndices[4 - 1].offset)

你应该检查数组的大小，而不是假设有一个像这样的最后一个值：

let values = [1, 3, 2, 4, 1, 3, 3, 4, 1]
let threesAndIndices = values.enumerated().filter { $1 == 3 }

// find index of last three
if let last = threesAndIndices.last {
    print("index of last three is \(last.offset)")
} else {
    print("there are no threes")
}

index of last three is 6

// find index of 4th three
let nth = 4
if nth > threesAndIndices.count {
    print("there aren't \(nth) threes")
} else {
    let index = threesAndIndices[nth - 1].offset
    print("the index of three #\(nth) is \(index)")
}

there aren't 4 threes

swift数组/集合中特定实例的索引

2 个答案: