从随机整数到实际字符串答案

【问题标题】：From random integers to an actual String从随机整数到实际字符串
【发布时间】：2011-09-29 02:36:32
【问题描述】：

这是一个包含 ASCII 表中所有打印字符的数组的代码。我正在尝试使整数形式的任何字符串消息（例如，转换为 97098097 的字符串“aba”可以放回其原始字符串形式。可以采用 100101101 并将其变回“dee”。我我真的很努力地尝试过这种方法，但它似乎不起作用，尤其是在涉及数字时，请帮助我。顺便说一下，它是用 Java 编写的，我正在使用 Eclipse。

public static String IntToString (){


int n = 0;
String message = "";
String message2 = null;
String [] ASCII = {" ","!","\"","#","$","%","&","\'","(",")","*","+",",","-",".","/","0","1","2","3","4","5","6","7","8","9",":",";","<","=",">","?","@","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","[","\\","]","^","_","`","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 IntMessage = result.toString();
String firstChar = IntMessage.substring(0,2);
if (IntMessage.substring(0,1)=="1" && IntMessage.length()%3==0)
{
    for (int x = (IntMessage.length() % 3 - 3) % 3; x < IntMessage.length()-2; x += 3)
        n = Integer.parseInt(IntMessage.substring(Math.max(x, 0), x + 3));
        message=message.concat(ASCII[n-31]);
return message;
}
else if (IntMessage.length()%3==2)
message2=ASCII[(Integer.parseInt(firstChar))-31];
        for (int x = 2; x < IntMessage.length()-2; x += 3)
            n = Integer.parseInt(IntMessage.substring(x, x + 3));
            message=message2+=ASCII [n - 31];
return message;

【问题讨论】：

嗯...为什么？你想用这个来达到什么目的？必须有一种更明智的方法来解决您的根本问题。
我知道我正在为一个项目做这个，但我无法让它工作我想要的只是：97098097=> "aba" 就是这样。
你能修复这段代码中的格式吗？您粘贴的标签会导致各种奇怪的缩进问题。目前您的代码示例不可读。此外，最初的 for 循环“太聪明了”，会导致这里的代码审查失败。

标签： java arrays string ascii integer

【解决方案1】：

您的编码方案似乎是，呃，疯狂。

首先，您获取字符串的 ASCII 值，然后获取该 ASCII 值的字符表示，然后将其存储为字符串。

所以"abc" => {97, 98, 99} => "979899".

但由于您使用的是 ASCII，它的值可以是 100 或更大，如果整数小于 100，则用 0 填充它们：

"abc" => {97, 98, 99} => {"097", "098", "099"} => "097098099"

但你决定只在某些时候这样做，因为不知何故

"aba" => "97098097"

即第一个“a”变成了“97”，而最后一个“a”变成了“097”。

我会说你应该先修复你的编码方案。

此外，希望这些不是“随机整数”，因为您正试图将它们变成合理的字符串。否则，像 base64 这样的简单映射很容易将任何整数映射到字符串，它们可能没有多大意义。

事实上，它们甚至不是真正的整数。您将编码字符串存储为字符串。

【讨论】：

【解决方案2】：

public static void main(String[] srgs){
    String aaa = "100101101";
    String[] a = split(aaa, 3);

    String s = "";

    for(int i=0;i<a.length;i++){
        char c = (char)Integer.parseInt(a[i]);
        s += Character.toString(c);
    }
    System.out.println(s);
}

public static String[] split(String str, int groupIndex){
    int strLength = str.length();
    int arrayLength = strLength/groupIndex;
    String[] splitedArray = new String[strLength/groupIndex];

    for(int i=0;i<arrayLength;i++){
        String splitedStr = str.substring(0, groupIndex);
        str = str.substring(groupIndex, str.length());
        arrayLength = str.length();
        splitedArray[i] = splitedStr;
    }
    return splitedArray;
}

最重要的是将ASCII字符串转换为Char值，而不是将其转换为字符串中的真实Character值。在这种情况下，需要将 ASCII 码长度固定为 3 会很有帮助。

【讨论】：