Question

我正致力于改变 Java文中的剖腹产。变化是,每2个空间都生成和插入随机信函。我的加密工作只是罚款。我的加密大多在发挥作用。我的方案对其加以加密,但我不知道如何去除随机信件。我写了撤销这些信件的职能,我目前的法典如下:

let decryptedMessage = seoekx tbhea mtidrniughkt

function finalMessage (decryptedMessage) {
    let finalMessage = "";
    for (let i = 1; i < decryptedMessage.length - 1; i++) {
        if (i % 2 === 0) {
            finalMessage += decryptedMessage.replace(decryptedMessage[i],"");
        }
        else {
            finalMessage += decryptedMessage[i];
        }
    }
    return finalMessage;
}

const finalMessageResult = finalMessage(decryptedMessageResult);
console.log(`The final decrypted message is ${finalMessageResult}`);

这是我的看法:

eseekx tbhea mtidrniughkteseoex tbhea mtidrniughktxseoekxtbhea mtidrniughkttseoekx thea mtidrniughkthsoekx tbhea mtidrniughktaseoekxtbhea mtidrniughktmseoekx bhea mtidrniughktiseoekx tbhea mtirniughktrseoekx tbhea mtidriughktiseoekx tbhea mtidrnighktgseoekx tbea mtidrniughktk

我认为,我需要把最后的婚姻从 lo中解放出来,但我不知道如何这样做,因为我感到,发言是否应该站在通过电文 lo上。

Any help would be appreciated.

我认为,我需要作一个发言。我头部的逻辑是使用2%=0,然后用“”取代这一要素。但显然,我所领导的以及正在执行的东西并不相同。我期望该守则将随机信件从扼杀中删除,从而形成可读的信息。

Answer 1

象你的调子一样,算术只是一小差一点,我们像你们一样,对算法进行过度思考! 引证:

let decryptedMessage = "seoekx tbhea mtidrniughkt";

function finalMessage (decryptedMessage) {
    let finalMessage = "";
    
    for (let i = 0; i < decryptedMessage.length; i++) {
        if ((i + 1) % 3 !== 0) {
            finalMessage += decryptedMessage[i];
        }
    }
    return finalMessage;
}

const finalMessageResult = finalMessage(decryptedMessage);
console.log(`The final decrypted message is ${finalMessageResult}`);

我做了一些改动:

Start your iterator at 0, because the first index of your string is 0
You want to mod 3, because you re removing every 3rd letter
Your algorithm was a bit unnecessarily complicated! This just adds every letter but the third letter to your output string
To compensate starting i at 0, I simply check i + 1

产出:<代码> 最终加密信息在午夜查询

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