将 5 位二进制字符串解码回字符

Question

我创建了一个简单的函数，它能够将每个字母字符 (a-z) 转换为 always 5-digit binary string (like "00000->a" or "11000->y")

SWIFT：

 let alphabeth_array=["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "", "", "", "", "", ""]
 for i in 0..<32 {
   let binary_code = String(i/16%2) + String(i/8%2) + String(i/4%2) + String(i/2%2) + String(i%2)
   print("\(alphabeth_array[i]) : \(binary_code)")
}

---

JAVASCRIPT：

 var alphabeth_array=["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "", "", "", "", "", ""]
 for (var i=0; i<32; i++) {
   var binary_code = (i/16%2).toString() + (i/8%2).toString() + (i/4%2).toString() + (i/2%2).toString() + (i%2).toString()
   console.log(alphabeth_array[i] + " : " + binary_code)
}

---

但是现在我想将二进制字符串解码回它们从上面的alphabeth_array 连接的字符。

SWIFT：

let binary = "00000" //starting with the binary string
let character = getChar(binaryCode: binary)

func getChar(binaryCode: String) -> (String) {
   // programm logic
   return result
}

---

JAVASCRIPT：

var binary = "00000" //starting with the binary string
var character = getChar(binary)


function getChar(binary) {
   // programm logic
   return result
}

我完全不知道如何反转解码函数以将二进制字符串转换回连接字符。任何帮助将不胜感激，谢谢！ （我宁愿只获得一些小代码 sn-ps、Swift 或 javascript - 哪种语言无关紧要 ！）

Answer 1

您可以使用init?<S>(_ text: S, radix: Int = default) where S : StringProtocol 初始化程序将二进制字符串转换为整数，并将其用作访问字母数组元素的索引。您也可以使用String(_ value:, radix:, uppercase:) 将数组索引转换为二进制的初始化程序：

let alphabeth = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"]
var bins: [String] = []
for index in alphabeth.indices {
    let binary = String(index, radix: 2)
    let zeros = repeatElement("0", count: 5 - binary.count).joined()
    bins.append("\(zeros)\(binary)")
    print(alphabeth[index], bins.last!)
}
for binary in bins {
    if let index = Int(binary, radix: 2) {
        print(alphabeth[index])
    }   
}

Answer 2

您可以使用基数2 解析字符串，添加10 的偏移量并将值转换为基数36 的值。

function getCharacter(string) {
    return (parseInt(string, 2) + 10).toString(36);
}

function getBinary(string) {
    return ('000000' + (parseInt(string, 36) - 10).toString(2)).slice(-5);
}

console.log(getCharacter('00000')); // a
console.log(getCharacter('11000')); // y
console.log(getBinary('a'));        // 00000
console.log(getBinary('y'));        // 11000

Answer 3

正如 Leo 在his answer 的第一句话中所说，可以将二进制字符串转换为 Swift 中的字符，如下所示：

let alphabetArray=["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "", "", "", "", "", ""]
let string = "00010"
let c = alphabetArray[Int(string, radix: 2)!]

或者你可以完全绕过alphabetArray：

let c = Character(Unicode.Scalar(Int(string, radix: 2)! + 97)!)

Answer 4

 const getChar = num => alphabeth_array[parseInt(num, 2)];