string - 将字符串拆分为特定长度的组

Question

如何在 Swift 中将给定的 String 分成具有给定长度的组，从右到左读取？

例如，我有字符串 123456789，组长度为 3。字符串应分为 3 组:123、456 , 789。字符串1234567会被分成1、234、567

那么，你能用 Swift 写一些漂亮的代码吗:

func splitedString(string: String, length: Int) -> [String] {

}

顺便说一句，尝试了函数 split()，但据我所知，它只适用于找到一些符号

Answer 1

swift 4

我改编了answer given by cafedeichi根据功能参数从左到右或从右到左操作，因此它更加通用。

extension String {
    /// Splits a string into groups of `every` n characters, grouping from left-to-right by default. If `backwards` is true, right-to-left.
    public func split(every: Int, backwards: Bool = false) -> [String] {
        var result = [String]()

        for i in stride(from: 0, to: self.count, by: every) {
            switch backwards {
            case true:
                let endIndex = self.index(self.endIndex, offsetBy: -i)
                let startIndex = self.index(endIndex, offsetBy: -every, limitedBy: self.startIndex) ?? self.startIndex
                result.insert(String(self[startIndex..<endIndex]), at: 0)
            case false:
                let startIndex = self.index(self.startIndex, offsetBy: i)
                let endIndex = self.index(startIndex, offsetBy: every, limitedBy: self.endIndex) ?? self.endIndex
                result.append(String(self[startIndex..<endIndex]))
            }
        }

        return result
    }
}

例子:

"abcde".split(every: 2)                     // ["ab", "cd", "e"]
"abcde".split(every: 2, backwards: true)    // ["a", "bc", "de"]

"abcde".split(every: 4)                     // ["abcd", "e"]
"abcde".split(every: 4, backwards: true)    // ["a", "bcde"]