Simple Encryption #1 - Alternating Split

题目

For building the encrypted string:
Take every 2nd char from the string, then the other chars, that are not every 2nd char, and concat them as new String.
Do this n times!

Examples:

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2
"This is a test!", 1 -> "hsi  etTi sats!"
"This is a test!", 2 -> "hsi  etTi sats!" -> "s eT ashi tist!"

Write two methods:

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2
function encrypt(text, n)
function decrypt(encryptedText, n)

For both methods:
If the input-string is null or empty return exactly this value!
If n is <= 0 then return the input text.

思路

正则大佬牛逼！
加密方案很简单，就是把字符串奇偶位抽出来再合并成一个新字符串。因为位置从0开始算，所以奇偶位是相反的。
解密方案：把加密后的字符串切成奇数串和偶数串，然后根据输入位的奇偶性分别从奇偶串把字符取出来，取完之后直接slice掉取完的字符。

答案

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function encrypt(text, n) {
    if (n < 1 || !text) {
        return text;
    } else {
        for (let i = 0; i < n; i++) {
            let str_even = "";
            let str_odd = "";
            for (let j = 0; j < text.length; j++) {
                (j % 2 === 0) ? (str_odd += text[j]) : (str_even += text[j]);
            }
            text = str_even + str_odd;
        }
        return text;
    }

}

function decrypt(encryptedText, n) {
    if (n < 1 || !encryptedText) {
        return encryptedText;
    } else {
        //构建单次解密函数
        function dec(enc_T) {
            let decryptText = "";
            //分割字符串
            let str_odd = enc_T.slice(enc_T.length / 2);
            let str_even = enc_T.slice(0, enc_T.length / 2);
            for (let i = 0; i < enc_T.length; i++) {
                if (i % 2 === 0) {
                    decryptText += str_odd[0];
                    str_odd = str_odd.slice(1);
                } else {
                    decryptText += str_even[0];
                    str_even = str_even.slice(1);
                }
            }
            return decryptText;
        };
        //判断解密次数
        for (let i = 0; i < n; i++) {
            encryptedText = dec(encryptedText);
        }
        return encryptedText;
    }
}