特定字符的Swift 4字符串范围

请注意,您的策略不适用于非嵌套括号。

E、 g.此输入

a(b)c(d)e

使用您的方法将产生此输出

ae

这是不正确的。

解决方案

此解决方案将与 嵌套的 和 非嵌套 括号

let text = "‌a(b)c(d)e"
var numOfNestedParentesis = 0
var indexesToRemove:Set<Int> = []

for (index, char) in text.enumerated() {

    if char == "(" {
        numOfNestedParentesis += 1
    }

    if numOfNestedParentesis > 0 {
        indexesToRemove.insert(index)
    }

    if char == ")" {
        numOfNestedParentesis -= 1
    }
}

let result = String(text.enumerated()
    .filter { !indexesToRemove.contains($0.offset) }
    .map { $0.element } )

测验

print(result) // ‌ace

这个 String.range 函数可以提供额外的选项,因此您可以通过 String.CompareOptions.backwards 获取最后一个实例的参数。

func deleteInnerString(with string: String, from firstCharacter: String, to secondCharacter: String) -> String {
    guard let leftIndex = (string.range(of: firstCharacter)?.lowerBound),
        let rightIndex = string.range(of: secondCharacter, options: .backwards)?.lowerBound else {
            return string
    }
    let remainingString = String(string.prefix(upTo: leftIndex) + string.suffix(from: string.index(after: rightIndex)))
    return remainingString
}

你必须使用 .backwards 选项 String.CompareOptions 在里面 range 字符串的函数。如下所示

func deleteString(with string: String, from firstCharacter: String, to secondCharacter: String) -> String {
    guard let leftIndex = (string.range(of: firstCharacter)?.lowerBound),
        let rightIndex = string.range(of: secondCharacter, options: .backwards)?.lowerBound else {
            return string
    }
    let remainingString = String(string.prefix(upTo: leftIndex) + string.suffix(from: string.index(after: rightIndex)))

    return remainingString.trimmingCharacters(in: .whitespaces)
}