摘要：由于我們得到摩斯密碼沒(méi)有空格隔開(kāi)，所以解密后有可能不止一種。完整的代碼得到結(jié)果如下，根據(jù)圖片中的提示，該單詞與面試有關(guān)，那么應(yīng)該是無(wú)疑。上面的代碼我們用了層嵌套循環(huán)，確實(shí)有點(diǎn)多，但是只有條件成立，才會(huì)進(jìn)入深層的循環(huán)。

今天在微信群里面看到一張招聘圖片，如下

被玷污的殘缺的婀娜多汁姿的二維碼，蘊(yùn)藏著通往神秘國(guó)度大門的鑰匙。利用微信自帶截圖工具，憑借著我手繪天賦，很快就還原了二維碼。

然后查看圖片，長(zhǎng)按。居然是可以識(shí)別的二維碼，掃一掃，得到下面的摩斯密碼

.--..-......

于是，百度摩斯密碼對(duì)照表。

由于我們得到摩斯密碼沒(méi)有空格隔開(kāi)，所以解密后有可能不止一種。這個(gè)時(shí)候，身為攻城虱當(dāng)然想到是用程序解決了。

既然，圖片中說(shuō)到是4個(gè)字的單詞，等于這段密碼可以分為4段。

[part1][part2][part3][part4]

如果4段密碼分別都能匹配上，那么連起來(lái)整個(gè)密碼也能匹配上（廢話）。最先想到是用4層循環(huán)窮舉法

for(){
    for(){
        for(){
            for(){
    
            }
        }
    }
}

可是如何把這段密碼分成四段呢？很簡(jiǎn)單，如果電碼符號(hào)與密碼前幾個(gè)字符能完全匹配，這里就分一段，然后繼續(xù)匹配下一段。

for(){
    // 匹配成功
    if([part1][part2][part3][part4].indexOf(xcode[i]) === 0 ){
        for(){
            if([part2][part3][part4].indexOf(xcode[j]) === 0 ){
                
            }
        }
    }
    
}

第四段，也就是最后一段匹配不能再用[indexOf]，而是用[===]。

核心匹配代碼

for(i = 0; i < morseCodeNum; i ++){
    if(mission1.indexOf(morseCodeMap[i]) === 0 ){
        firstMatchPart = morseCodeMap[i];
        mission2 = secretCode.slice(firstMatchPart.length);

        for(j = 0; j < morseCodeNum; j ++){

            if(mission2.indexOf(morseCodeMap[j]) === 0 ){
                secondMatchPart = morseCodeMap[j];
                mission3 = secretCode.slice(firstMatchPart.length+secondMatchPart.length);

                for(k = 0; k < morseCodeNum; k ++){
                    if(mission3.indexOf(morseCodeMap[k]) === 0 ){
                        thirdMatchPart = morseCodeMap[k];
                        mission4 = secretCode.slice(firstMatchPart.length+secondMatchPart.length+thirdMatchPart.length);

                        for(n = 0; n < morseCodeNum; n ++){
                            if(mission4 === morseCodeMap[n]){
                                fourthMatchPart = morseCodeMap[n];
                                console.log([firstMatchPart, secondMatchPart, thirdMatchPart, fourthMatchPart].join(""));
                                result.push([letters[i], letters[j], letters[k], letters[n]].join(""));
                                }
                            }
                        }
                    }
                }
            }
        }
    }

注: mission1表示處于匹配第一段密碼，以此類推。

完整的代碼

var morseCodeMap = [
    ".-", "-...", "-.-.", "-..",
    ".",  "..-.", "--.", "....",
    "..", ".---", "-.-", ".-..",
    "--", "-.", "---", ".--.",
    "--.-", ".-.", "...", "-",
    "..-", "...-", ".--", "-..-",
    "-.--", "--.."
], 
secretCode = ".--..-......",
morseCodeNum = morseCodeMap.length,
letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"; 

var result = [],
    firstMatchPart,
    secondMatchPart,
    thirdMatchPart,
    fourthMatchPart,
    mission1,
    mission2,
    mission3,
    mission4;

function deCodeMorse(){
    var i, j, k, n;
    mission1 = secretCode;
    for(i = 0; i < morseCodeNum; i ++){
        if(mission1.indexOf(morseCodeMap[i]) === 0 ){
            firstMatchPart = morseCodeMap[i];
            mission2 = secretCode.slice(firstMatchPart.length);

            for(j = 0; j < morseCodeNum; j ++){

                if(mission2.indexOf(morseCodeMap[j]) === 0 ){
                    secondMatchPart = morseCodeMap[j];
                    mission3 = secretCode.slice(firstMatchPart.length+secondMatchPart.length);

                    for(k = 0; k < morseCodeNum; k ++){
                        if(mission3.indexOf(morseCodeMap[k]) === 0 ){
                            thirdMatchPart = morseCodeMap[k];
                            mission4 = secretCode.slice(firstMatchPart.length+secondMatchPart.length+thirdMatchPart.length);

                            for(n = 0; n < morseCodeNum; n ++){
                                if(mission4 === morseCodeMap[n]){
                                    fourthMatchPart = morseCodeMap[n];
                                    console.log([firstMatchPart, secondMatchPart, thirdMatchPart, fourthMatchPart].join(""));
                                    result.push([letters[i], letters[j], letters[k], letters[n]].join(""));
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}

deCodeMorse();
console.log(result);    

得到結(jié)果如下，根據(jù)圖片中的提示，該單詞與面試有關(guān)，那么應(yīng)該是pass無(wú)疑。

上面的代碼我們用了4層for嵌套循環(huán)，確實(shí)有點(diǎn)多，但是只有if條件成立，才會(huì)進(jìn)入深層的循環(huán)。因?yàn)橐笏薪?，是避免不了?

假如，我們得到的密碼可能不是4段，不確定是幾段，這個(gè)時(shí)候就不能用嵌套for循環(huán)了，可以用遞歸。
核心代碼邏輯

function deCodeMorse(mission[j]){
    for (var i = 0; i < morseCodeNum; i++) {
        if( morseCode.indexOf(morseCodeMap[i]) === 0 ){
            if( morseCode === morseCodeMap[i]){
                // 輸出匹配結(jié)果
            }else{
                deCodeMorse(mission[j+1]);
            }
        }
    }
}

使用遞歸的完整代碼

var morseCodeMap = [
        ".-", "-...", "-.-.", "-..",
        ".",  "..-.", "--.", "....",
        "..", ".---", "-.-", ".-..",
        "--", "-.", "---", ".--.",
        "--.-", ".-.", "...", "-",
        "..-", "...-", ".--", "-..-",
        "-.--", "--.."
    ], 
    secretCode = ".--..-......",
    morseCodeNum = morseCodeMap.length,
    letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ",
    result = [],


function deCodeMorse(morseCode, deCodeResult, stack){
    stack = stack || [];
    for (var i = 0; i < morseCodeNum; i++) {
        if( morseCode.indexOf(morseCodeMap[i]) === 0 ){
            if( morseCode === morseCodeMap[i]){
                deCodeResult.push(stack.concat(letters[i]).join(""));
            }else{
                deCodeMorse(morseCode.slice(morseCodeMap[i].length), deCodeResult, stack.concat(letters[i]));
            }
        }
    }
}

deCodeMorse(secretCode, result);
console.log(result);

我們還可以傳一個(gè)數(shù)字參數(shù)給函數(shù)deCodeMorse，表示密碼由幾段組成，不傳則沒(méi)有限制。那么deCodeMorse函數(shù)可以這樣寫

function deCodeMorse(morseCode, deCodeResult, limit, stack){
    stack = stack || [];
    for (var i = 0; i < morseCodeNum; i++) {
        if( morseCode.indexOf(morseCodeMap[i]) === 0 ){
            if( morseCode === morseCodeMap[i]){
                if(limit){
                    stack.length +1 === limit && deCodeResult.push(stack.concat(letters[i]).join(""));
                }else{
                    deCodeResult.push(stack.concat(letters[i]).join(""));
                }
            }else{
                if(limit){
                    if(stack.length < limit){
                        deCodeMorse(morseCode.slice(morseCodeMap[i].length), deCodeResult, limit, stack.concat(letters[i]));
                    }
                }else{
                    deCodeMorse(morseCode.slice(morseCodeMap[i].length), deCodeResult, limit, stack.concat(letters[i]));
                }
                
            }
        }
    }
}

deCodeMorse(secretCode, result, 4);