Question

我是JS的初学者。我试图了解Caesar Cipher ROT13，但对我来说太复杂了。因此，我尝试编写自己的代码。在下面：

public class Home implements EasyPermissions.PermissionCallbacks{

...
private String[] permission_sounds = {Manifest.permission.WRITE_SETTINGS, Manifest.permission.WRITE_EXTERNAL_STORAGE};
private int RC_permissionSounds = 500;
private Activity act;

..onCreate{
    act = this;
}

//I assign this onClickListener to ringtoneButton
private View.OnClickListener ringtoneMenuOpen(){
        Log.d(TAG, "ringtoneMenuOpen accedemos, tenemos que entrar ahora en solicitar permisos e intentar ");
        soundButtonMode = SoundButtonMode.RIGNTONE;
        return new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                Log.d(TAG, "He pulsado ringtoneMenuOpen"); //Line 1 Logcat
                if (EasyPermissions.hasPermissions(act, permission_sounds)) {
                    new SetSoundAsRingtone(act).execute();
                } else {
                    EasyPermissions.requestPermissions(act, act.getString(R.string.permission_rationale), RC_permissionSounds, permission_sounds);
                }
            }
        };
    }
...
    @Override
    public void onRequestPermissionsResult(int requestCode, String[] permissions, int[] grantResults) {
        super.onRequestPermissionsResult(requestCode, permissions, grantResults);
        EasyPermissions.onRequestPermissionsResult(requestCode, permissions, grantResults, act);
    }
    @Override
    public void onPermissionsGranted(int requestCode, @NonNull List<String> perms) {
        Log.d(TAG, "onPermissionsGranted");
        buttonRingtone.performClick();
    }
    @Override
    public void onPermissionsDenied(int requestCode, @NonNull List<String> perms) {
        Log.d(TAG, "onPermissionsDenied");//Line 2 Logcat

        if (EasyPermissions.somePermissionPermanentlyDenied(this, perms)) {
            Log.d(TAG, "Hemos llegado aquí con permisos denegados y 'No volver a mostrar'"); //Line 3 Logcat
            new AppSettingsDialog.Builder(this).build().show();
        }
    }
...

但是我被困住了。怎么做：

1。。从var str中获取值，然后访问var字母，将每个字母从var str值更改为字母表中的下3个字母（var str，每个元素的当前位置都会更改）。：function encrip() { var alphabet = ["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 str = "Ni Hao"; var string = str.toUpperCase(); for (var i = 0; i < string.length; i++) { for (var k = 0; k < alphabet.length; k++) { if(string.charAt(i) == alphabet[k]) { /* console.log(string.charAt(i) + ' ' + alphabet.indexOf(alphabet[k])); */ } } } } encrip();

2。。我的意思是，创建通用代码，不仅是为了将位置更改3，而且当我给定值“ 5”时，每个元素的字母顺序都将被更改5个位置。因此，当我更改其值时可以更改输出

我希望我能清楚地解释一切。预先感谢大家的帮助！

Answer 1

您可以使用以下函数对英语单词进行加密，第一个参数是要加密的字符串，第二个参数是要移位的

function encryp(str,pos){
   var alpha="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
   var strUC=str.toUpperCase();
   var enc="";
    for(var i=0;i<strUC.length;i++){
    if(strUC.charAt(i)!=" "){
    enc+=alpha.charAt((alpha.indexOf(strUC.charAt(i))+pos)%26)
     }
  else{
   enc+=" "
     }
   // in your case pos=3
    }
    
    return enc;
    }
console.log(encryp("NiHao",3));

Answer 2

您不需要两个for循环即可执行此操作。遍历输入string并在alphabet数组中找到每个字符的索引，如果找到，则向其中添加shift以得到加密的字符。
要处理溢出，请使用模运算符在数组中循环。我还假设您不会使用任何特殊符号进行加密。

function encrip(string, shift) {
  var alphabet = ["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"];
  string = string.toUpperCase();
  let arr = [];
    for (var i = 0; i < string.length; i++) {
       let char = alphabet.indexOf(string[i]) !== -1 ? alphabet[(alphabet.indexOf(string[i]) %26) + shift] : " ";
       arr.push(char);
    }
  let encryp = arr.join("");
  console.log(encryp);
  return encryp;
}
encrip("Ni Hao", 3);

Answer 3

首先，可以使用内置函数indexOf：

，而不是内部for循环扫描整个alphabet数组：

alphabet.indexOf('K') // returns 10

第二，您需要在单独的变量中构建加密的字符串。对于每个字母，获取该字母在字母表中的索引，将密码偏移量参数添加到该索引，并将结果从字母表中添加到新字符串中。重要的一步是，当添加到字母的索引时，您要确保结果索引在alphabet数组的范围内。您可以使用%（模）运算符来完成此操作，该运算符会将高值回绕到数组的开头。全部：

function encipher(input, offset) {
    var alphabet = ["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 str = input.toUpperCase();
    var result = '';

    for (var i = 0; i < str.length; i++) {
        letterIndex = alphabet.indexOf(str.charAt(i));
        if (letterIndex === -1) {
            result += str[i]; // if the letter isn't found in the alphabet, add it to the result unchanged
            continue;
        }

        cipheredIndex = (letterIndex + offset) % alphabet.length; // wrap index to length of alphabet
        result += alphabet[cipheredIndex];
    }

    console.log(result);
}

encipher('Ni Hao', 5); // output: 'SN MFT'

如何使数组中的元素更改其位置

3 个答案: