Question

我想在Swift中将字符串拆分为两个符号。所以在字符串“df57g5df7g”之后我想获得一个数组[“df”，“57”，“g5”，“df”，“7g”]。是否可以强制使用迭代器

for i in word.characters {
 print(i)
}

跳过两个符号，并获得循环中的下一个符号？

Answer 1

一个简单的while循环：

let str = "df57g5df7g"

var startIndex = str.startIndex
var result = [String]()

repeat {
    let endIndex = startIndex.advancedBy(2, limit: str.endIndex)
    result.append(str[startIndex..<endIndex])

    startIndex = endIndex
} while startIndex < str.endIndex

print(result)

或者更多的东西：

let result = 0.stride(to: str.characters.count, by: 2).map { i -> String in
    let startIndex = str.startIndex.advancedBy(i)
    let endIndex   = startIndex.advancedBy(2, limit: str.endIndex)
    return str[startIndex..<endIndex]
}

Answer 2

这可能不是最简单的解决方案，但它有效：

var word = "df57g5df7g"

var pairsArray = [String]()

while word.characters.count > 1 {
    let firstCharacter = word.removeAtIndex(word.startIndex)
    let secondCharacter = word.removeAtIndex(word.startIndex)
    pairsArray.append("\(firstCharacter)\(secondCharacter)")
}

print(pairsArray)

结果是：

["df", "57", "g5", "df", "7g"]

Answer 3

这是我见过的最佳解决方案，取自SwiftSequence library。

extension CollectionType {
    public func chunk(n: Index.Distance) -> [SubSequence] {
        var res: [SubSequence] = []
        var i = startIndex
        var j: Index
        while i != endIndex {
            j = i.advancedBy(n, limit: endIndex)
            res.append(self[i..<j])
            i = j
        }
        return res
    }
}

let word = "df57g5df7g"
let pairs = word.characters.chunk(2).map(String.init)
print(pairs) //["df", "57", "g5", "df", "7g"]

您可以在行动here中看到它。

在Swift中用两个符号拆分字符串

3 个答案: