在Swift中拆分字符串而不删除定界符

此方法适用于

CollectionTypes

String

extension CollectionType {  func splitAt(@noescape isSplit: Generator.Element throws -> Bool) rethrows ->  [SubSequence] {    var p = startIndex    return try indices      .filter { i in try isSplit(self[i]) }      .map { i in        defer { p = i }        return self[p..<i]      } + [suffixFrom(p)]  }}extension CollectionType where Generator.Element : Equatable {  func splitAt(splitter: Generator.Element) -> [SubSequence] {    return splitAt { el in el == splitter }  }}

您可以这样使用它：

let sentence = "Hello, my name is oisdk. This should split: but only at punctuation!"let puncSet = Set("!.,:".characters)sentence  .characters  .splitAt(puncSet.contains)  .map(String.init)// ["Hello", ", my name is oisdk", ". This should split", ": but only at punctuation", "!"]

或者，此版本使用for循环，并 在 定界符 后 拆分：

extension CollectionType {  func splitAt(@noescape isSplit: Generator.Element throws -> Bool) rethrows ->  [SubSequence] {    var p = startIndex    var result: [SubSequence] = []    for i in indices where try isSplit(self[i]) {      result.append(self[p...i])      p = i.successor()    }    if p != endIndex { result.append(suffixFrom(p)) }    return result  }}extension CollectionType where Generator.Element : Equatable {  func splitAt(splitter: Generator.Element) -> [SubSequence] {    return splitAt { el in el == splitter }  }}let sentence = "Hello, my name is oisdk. This should split: but only at punctuation!"let puncSet = Set("!.,:".characters)sentence  .characters  .splitAt(puncSet.contains)  .map(String.init)// ["Hello,", " my name is oisdk.", " This should split:", " but only at punctuation!"]

或者，如果你想获得最斯威夫特功能为一体的功能（

defer

throws

flatMap

guard

extension CollectionType {  func splitAt(@noescape isSplit: Generator.Element throws -> Bool) rethrows -> [SubSequence] {    var p = startIndex    var result: [SubSequence] = try indices.flatMap { i in      guard try isSplit(self[i]) else { return nil }      defer { p = i.successor() }      return self[p...i]    }    if p != endIndex { result.append(suffixFrom(p)) }    return result  }}