blocks|key|4337364|text|按数字迭代数字数字，并使用开关-大小写子句来决定必须映射到哪个字符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4337365|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Iterate through the number digit by digit and use switch-case clauses to decide which character you have to map it to.

blocks|key|4356904|text|提示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4356905|模10表示最后一位数
除以整数的10除去它的最后一个数字。|blockquote|4356906|4356907|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|I|8|@]|9|@]|A|$]]|$1|E|3|-4|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Hint :
<blockquote>
Modulo by 10 gives last digit
Division by 10 of an integer removes its last digit.
</blockquote>

blocks|key|4356933|text|这是算法的一般思想，修改它以创建char数组，而不是打印结果：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4356934|void+printNumber(int+n)+{
++++if+(n+<+10)
++++++++System.out.println(n);
++++else+{
++++++++printNumber(n+/+10);
++++++++System.out.println(n+%25+10);
++++}
}|code-block|syntax|javascript|4356935|我用递归算法写的，因为我喜欢递归.但是将其转换为迭代解决方案是很简单的。|4356936|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is the general idea of the algorithm, modify it to create a char array instead of printing the results:

<pre><code>void printNumber(int n) {
 if (n &lt; 10)
 System.out.println(n);
 else {
 printNumber(n / 10);
 System.out.println(n % 10);
 }
}
</code></pre>

I wrote it using a recursive algorithm, because I fancy recursion ... but it's trivial to convert this to an iterative solution.

blocks|key|4337455|text|public+static+void+main(String[]+args)+{
++++final+int+number+=+500;
++++final+int+base+=+10;+//+the+number+system,+in+your+case:+decimal
++++int+tmp+=+number;+//+create+a+temporary+variable+for+processing
++++StringBuilder+sb+=+new+StringBuilder();+//+create+a+buffer+for+your+textual+representation+(or+calc+the+length+beforehand)
++++//
++++do+{
++++++++int+digit+=+tmp+%25+base;+//+get+the+last+digit
++++++++sb.append(digit);+//+attach+it+to+your+character+buffer
++++++++tmp+/=+base;+//+remove+the+numbers+last+digit
++++}+while+(tmp+!=+0);+//+stop+when+there+is+no+more+character+but+when+the+number+is+0,+you+have+to+execute+it+once
++++sb.reverse();+//+now+flip+the+characters+because+it+was+written+to+from+behind
++++//
++++String+numText+=+sb.toString();
++++System.out.println(numText);
++++//
++++char[]+chars+=+numText.toCharArray();
++++sb+=+new+StringBuilder("{");
++++for+(int+i+=+0;+i+<+chars.length;+i%2B%2B)+{
++++++++char+c+=+chars[i];
++++++++sb.append(c);
++++++++if+(i+!=+chars.length+-+1)+{
++++++++++++sb.append(",+");
++++++++}
++++}
++++sb.append("}");
++++//
++++System.out.println(sb.toString());
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4337456|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public static void main(String[] args) {
 final int number = 500;
 final int base = 10; // the number system, in your case: decimal
 int tmp = number; // create a temporary variable for processing
 StringBuilder sb = new StringBuilder(); // create a buffer for your textual representation (or calc the length beforehand)
 //
 do {
 int digit = tmp % base; // get the last digit
 sb.append(digit); // attach it to your character buffer
 tmp /= base; // remove the numbers last digit
 } while (tmp != 0); // stop when there is no more character but when the number is 0, you have to execute it once
 sb.reverse(); // now flip the characters because it was written to from behind
 //
 String numText = sb.toString();
 System.out.println(numText);
 //
 char[] chars = numText.toCharArray();
 sb = new StringBuilder("{");
 for (int i = 0; i &lt; chars.length; i++) {
 char c = chars[i];
 sb.append(c);
 if (i != chars.length - 1) {
 sb.append(", ");
 }
 }
 sb.append("}");
 //
 System.out.println(sb.toString());
}
</code></pre>

blocks|key|4356977|text|如果我们完全遵循这些限制，这是一个相当不错的练习。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4356978|这是我的解决方案：|4356979|static+char[]+getCharArrayFromInt(int+value)+{
++++if+(value+==+0)
++++++++return+new+char[]{'0'};
++++if+(value+==+Integer.MIN_VALUE)
++++++++return+new+char[]{'-',+'2',+'1',+'4',+'7',+'4',+'8',+'3',+'6',+'4',+'8'};
++++int+signValue+=+0;
++++if+(value+<+0)+{
++++++++signValue+=+1;
++++++++value+*=+-1;
++++}
++++char[]+result+=+new+char[signValue+%2B+(int)+(Math.log10(value)+%2B+1)];
++++//+to+be+correct,+we+need+a+switch/function+here.+but+for+the+length,+lets+reduce+it
++++if+(signValue+==+1)
++++++++result[0]+=+'-';
++++for+(int+i+=+result.length+-+1;+i+>=+0+%2B+signValue;+--i)+{
++++++++result[i]+=+(char)+((value+%25+10)+%2B+'0');
++++++++//+to+be+correct,+we+need+a+switch+here.+but+for+the+length,+lets+reduce+it
++++++++value+/=+10;
++++}
++++return+result;
}|code-block|syntax|javascript|4356980|它违反了源中标记的“不使用任何内置的java转换函数”两次。这只是为了缩短计划。为这个部分编写一个带有开关的函数是非常清楚和容易的。|4356981|函数应该对完整的整数范围有效。|4356982|但老实说，我不认为一个人能在10分钟的工作面试中解决这个问题。您应该开始讨论如何编写最干净的代码，以及为什么最好避免这样的事情，如上面所示。|4356983|-结核病|4356984|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

If we follow the restrictions completely, this is a rather nice exercise.

Here is my solution:

<pre><code>static char[] getCharArrayFromInt(int value) {
 if (value == 0)
 return new char[]{'0'};
 if (value == Integer.MIN_VALUE)
 return new char[]{'-', '2', '1', '4', '7', '4', '8', '3', '6', '4', '8'};
 int signValue = 0;
 if (value &lt; 0) {
 signValue = 1;
 value *= -1;
 }
 char[] result = new char[signValue + (int) (Math.log10(value) + 1)];
 // to be correct, we need a switch/function here. but for the length, lets reduce it
 if (signValue == 1)
 result[0] = '-';
 for (int i = result.length - 1; i &gt;= 0 + signValue; --i) {
 result[i] = (char) ((value % 10) + '0');
 // to be correct, we need a switch here. but for the length, lets reduce it
 value /= 10;
 }
 return result;
}
</code></pre>

It violates the "without using any inbuilt java conversion function" two times, marked in the source. This was only to shorten the program. It is very clear and easy to write a function with a switch for this parts.

The functions should be valid for the complete Integer range.

But honestly, I do not think one could make this solution in 10 min in a job interview. You should start discussing about writing clean code pays off the most and why is it better to avoid such things as shown above.

--tb

I am preparing for an interview and I am not able to solve one of the problem.
Below is the problem:

<blockquote>
 Take number and print it as character one by one without using any inbuilt java conversion function.
 Below is the example lets say you have one numnber 1234 and now write function to take this number as integer and print character array {1,2,3,4}.
</blockquote>

It asked to write this function without using any inbuilt java conversion
Please let me if any one knows how to solve it.

How to convert number into character array in java without using any java inbuilt function

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在为面试做准备，我无法解决其中的一个问题。以下是问题所在：取数字作为字符逐个打印，不使用任何内置的java转换函数。下面是一个示例，假设您有一个numnber 1234，现在使用这个数字作为整数并打印字符数组{1,2,3,4}。它要求在不使用内置的java转换的情况下编写这个函数，如果有人知道如何解决它，请让我来。

问如何在java中不使用任何java内置函数将数字转换为字符数组
EN

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问如何在java中不使用任何java内置函数将数字转换为字符数组EN