快速向后搜索数组

Question

我有一个数组，我想向后搜索。我对原始数组的结果索引感兴趣，而不是实际值。我如何将ReverseRandomAccessIndex转换为“正常”索引？

let arr = [1,2,3,4,5] 
if let index = arr.reverse().indexOf(2) //index is ReverseRandomAccessIndex 
{ 
let searchvalue = arr.reverse()[index] 
} 

Answer 1

很简单：

let arr = [1, 2, 3, 4, 5] 

if let index = arr.reverse().indexOf(2) 
{ 
    let searchvalue = arr[index.base - 1] 
} 

base，如在标准库中定义：

/// The successor position in the underlying (un-reversed) 
    /// collection. 
    /// 
    /// If `self` is `advance(c.reverse.startIndex, n)`, then: 
    /// - `self.base` is `advance(c.endIndex, -n)`. 
    /// - if `n` != `c.count`, then `c.reverse[self]` is 
    /// equivalent to `[self.base.predecessor()]`. 
    public let base: Base