ios - 如何按给定大小的 block 从字符串拆分为数组

Question

我想按给定大小的 block 拆分字符串 2

示例:

字符串 "1234567" 和输出应该是 ["12", "34", "56","7"]

Answer 1

您可以按照以下每 n 个元素对您的集合元素(在本例中为字符)进行分组:

extension Collection {
    func unfoldSubSequences(limitedTo maxLength: Int) -> UnfoldSequence<SubSequence,Index> {
        sequence(state: startIndex) { start in
            guard start < self.endIndex else { return nil }
            let end = self.index(start, offsetBy: maxLength, limitedBy: self.endIndex) ?? self.endIndex
            defer { start = end }
            return self[start..<end]
        }
    }
    func subSequences(of n: Int) -> [SubSequence] {
        .init(unfoldSubSequences(limitedTo: n))
    }
}

---

let numbers = "1234567"
let subSequences = numbers.subSequences(of: 2)
print(subSequences)    // ["12", "34", "56", "7"]

编辑/更新:

如果您想将超出的字符附加到最后一组:

extension Collection {
    func unfoldSubSequencesWithTail(lenght: Int) -> UnfoldSequence<SubSequence,Index> {
        let n = count / lenght
        var counter = 0
        return sequence(state: startIndex) { start in
            guard start < endIndex else { return nil }
            let end = index(start, offsetBy: lenght, limitedBy: endIndex) ?? endIndex
            counter += 1
            if counter == n {
                defer { start = endIndex }
                return self[start...]
            } else {
                defer { start = end }
                return self[start..<end]
            }
        }
    }
    func subSequencesWithTail(n: Int) -> [SubSequence] {
        .init(unfoldSubSequencesWithTail(lenght: n))
    }
}

---

let numbers = "1234567"
let subSequencesWithTail = numbers.subSequencesWithTail(n: 2)
print(subSequencesWithTail)    // ["12", "34", "567"]