枚举swift字符串时获取“string.index”

Question

let greeting = "Hello"
for (intIndex, char) in greeting.enumerated() {
    let currentIndex = greeting.index(greeting.startIndex, offsetBy: intIndex)
    let indexAfterCurrentIndex = greeting.index(after: currentIndex)
    print(greeting[indexAfterCurrentIndex...])
}

let currentIndex = greeting.index(greeting.startIndex, offsetBy: intIndex)

let greeting = "Hello"
for (stringIndex, char) in greeting.enumeratedXXX() {
    let indexAfterCurrentIndex = greeting.index(after: stringIndex)
    print(greeting[indexAfterCurrentIndex...])
}

Answer 1

对此没有内置功能。您可以将其封装在自定义迭代器中，但您只能将相同类型的计算封装在不同的地方，所以这不是答案：）

但是，您可以改进当前代码的性能：

greeting.index(greeting.startIndex, offsetBy: intIndex)

• 这将为每个循环迭代计算从startindex到结果索引的索引。
• 使用索引(_:offsetby:)的索引计算实际上只是另一个循环本身，其中每个索引+1。没有O(1)方法来“计算”索引；它是通过O(n)
中的循环找到的

您可以将复杂性降低到每次迭代2个操作，或者O(2n)。只需将以前计算的索引保存在内存中并+1自己，避免从头开始重新计算。

代码如下：

let greeting = "Hello"
var index = greeting.startIndex
for char in greeting {
    let indexAfterCurrentIndex = greeting.index(after: index)
    print(greeting[indexAfterCurrentIndex...])
    index = indexAfterCurrentIndex
}

仍然不是一个简单和内置的解决方案，但你也可以包装这个更有效的算法，然后离开！

extension String {
    func forEachCharacterWithIndex(iterator: (String.Index, Character) -> Void) {
        var currIndex = self.startIndex
        for char in self {
            iterator(currIndex, char)
            currIndex = self.index(after: currIndex)
        }
    }
}

let greeting = "Hello"
greeting.forEachCharacterWithIndex { (index, char) in
    let indexAfterCurrentIndex = greeting.index(after: index)
    print(greeting[indexAfterCurrentIndex...])
}