如何转换字符串？

下面是一个解决这个问题的例子，我以ipatev_nn提出的正则表达式为基础，给它加上了分组的括号，虽然这个正则表达式只适用于form的字符串，4{b3{a}}不适合嵌套较多的字符串组（例如），您可以在此处4{b3{a6{c}}}通过解析检查正则表达式

        let str = NSString("4{b3{a}}")
        let pattern = "^([0-9]+)\\{([a-z]+)([0-9]+)\\{([a-z]+)\\}\\}$"
        let regex = try! NSRegularExpression(pattern: pattern, options: [])
        
        let matches = regex.matches(in: str as String, options: [], range: NSRange(location: 0, length: str.length))
        var values = [String]()
        
        for match in matches {
            for rangeIndex in 1..<match.numberOfRanges {
                let range = match.range(at: rangeIndex)
                let value = str.substring(with: range)
                print(range, value)
                values.append(value)
            }
        }

结果，我们得到这样一个数组，其中偶数位置被要重复的子串占据，奇数位置是重复的次数

4
b
3
a

我们基于数组生成一个字符串，从内部组开始（从数组末尾开始），也从末尾逐渐形成字符串，每次在开头插入新数据

        var result = ""
        
        for i in stride(from: values.count - 2, through: 0, by: -2) {
            let repeatCount = Int(values[i])!
            let value = values[i + 1]
            result.insert(contentsOf: value, at: result.startIndex)
            result = String(repeating: result, count: repeatCount)
        }

完整示例

    func buildString() {
        
        // parse encoded string
        
        let str = NSString("4{b3{a}}")
        let pattern = "^([0-9]+)\\{([a-z]+)([0-9]+)\\{([a-z]+)\\}\\}$"
        let regex = try! NSRegularExpression(pattern: pattern, options: [])
        
        let matches = regex.matches(in: str as String, options: [], range: NSRange(location: 0, length: str.length))
        var values = [String]()
        
        for match in matches {
            for rangeIndex in 1..<match.numberOfRanges {
                let range = match.range(at: rangeIndex)
                let value = str.substring(with: range)
                print(range, value)
                values.append(value)
            }
        }
        
        // build result string
        
        var result = ""
        
        for i in stride(from: values.count - 2, through: 0, by: -2) {
            let repeatCount = Int(values[i])!
            let value = values[i + 1]
            result.insert(contentsOf: value, at: result.startIndex)
            result = String(repeating: result, count: repeatCount)
        }
        
        print(result)
    }