blocks|key|1182949|text|将其转换为String并使用String#toCharArray()或String#split()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1182950|String+number+=+String.valueOf(someInt);

char[]+digits1+=+number.toCharArray();
//+or:
String[]+digits2+=+number.split("(?<=.)");|code-block|syntax|javascript|1182951|如果您已经在使用Java8，然后碰巧想要对其执行一些聚合操作，请考虑使用String#chars()从它中提取一个IntStream。|1182952|IntStream+chars+=+number.chars();|1182953|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/8/docs/api/java/lang/String.html|1|http://docs.oracle.com/javase/8/docs/api/java/lang/String.html#toCharArray--|2|http://docs.oracle.com/javase/8/docs/api/java/lang/String.html#split-java.lang.String-|3|http://docs.oracle.com/javase/8/docs/api/java/lang/CharSequence.html#chars--|4|http://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html^0|5|6|E|K|Z|E|5|6|0|E|K|1|Z|E|2|0|0|10|E|1L|9|10|E|3|1L|9|4|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@$9|1B|A|1C|1|1D]|$9|1E|A|1F|1|1G]|$9|1H|A|1I|1|1J]]|E|$]]|$1|F|3|G|5|H|7|1K|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1L|8|@$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]]|D|@$9|1Q|A|1R|1|1S]|$9|1T|A|1U|1|1V]]|E|$]]|$1|M|3|N|5|H|7|1W|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1X|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]|10|$5|R|S|T|E|$U|11]]|12|$5|R|S|T|E|$U|13]]]]

Convert it to <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/String.html" rel="noreferrer"><code>String</code></a> and use <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/String.html#toCharArray--" rel="noreferrer"><code>String#toCharArray()</code></a> or <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/String.html#split-java.lang.String-" rel="noreferrer"><code>String#split()</code></a>.

<pre><code>String number = String.valueOf(someInt);

char[] digits1 = number.toCharArray();
// or:
String[] digits2 = number.split("(?&lt;=.)");
</code></pre>

In case you're already on Java 8 and you happen to want to do some aggregate operations on it afterwards, consider using <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/CharSequence.html#chars--" rel="noreferrer"><code>String#chars()</code></a> to get an <a href="http://docs.oracle.com/javase/8/docs/api/java/util/stream/IntStream.html" rel="noreferrer"><code>IntStream</code></a> out of it.

<pre><code>IntStream chars = number.chars();
</code></pre>

blocks|key|968683|text|这个怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|968684|public+static+void+printDigits(int+num)+{
++++if(num+/+10+>+0)+{
++++++++printDigits(num+/+10);
++++}
++++System.out.printf("%25d+",+num+%25+10);
}|code-block|syntax|javascript|968685|或者，我们可以将其收集到一个整数数组中，然后打印该数组，而不是打印到控制台：|968686|public+static+void+main(String[]+args)+{
++++Integer[]+digits+=+getDigits(12345);
++++System.out.println(Arrays.toString(digits));
}

public+static+Integer[]+getDigits(int+num)+{
++++List<Integer>+digits+=+new+ArrayList<Integer>();
++++collectDigits(num,+digits);
++++return+digits.toArray(new+Integer[]{});
}

private+static+void+collectDigits(int+num,+List<Integer>+digits)+{
++++if(num+/+10+>+0)+{
++++++++collectDigits(num+/+10,+digits);
++++}
++++digits.add(num+%25+10);
}|968687|如果您希望保持数字的顺序，从最低有效位(索引)到最高有效位(索引n)，则需要使用以下更新后的getDigits()：|968688|/**
+*+split+an+integer+into+its+individual+digits
+*+NOTE:+digits+order+is+maintained+-+i.e.+Least+significant+digit+is+at+index[0]
+*+@param+num+positive+integer
+*+@return+array+of+digits
+*/
public+static+Integer[]+getDigits(int+num)+{
++++if+(num+<+0)+{+return+new+Integer[0];+}
++++List<Integer>+digits+=+new+ArrayList<Integer>();
++++collectDigits(num,+digits);
++++Collections.reverse(digits);
++++return+digits.toArray(new+Integer[]{});
}|968689|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

How about this?

<pre><code>public static void printDigits(int num) {
 if(num / 10 &gt; 0) {
 printDigits(num / 10);
 }
 System.out.printf("%d ", num % 10);
}
</code></pre>

or instead of printing to the console, we can collect it in an array of integers and then print the array:

<pre><code>public static void main(String[] args) {
 Integer[] digits = getDigits(12345);
 System.out.println(Arrays.toString(digits));
}

public static Integer[] getDigits(int num) {
 List&lt;Integer&gt; digits = new ArrayList&lt;Integer&gt;();
 collectDigits(num, digits);
 return digits.toArray(new Integer[]{});
}

private static void collectDigits(int num, List&lt;Integer&gt; digits) {
 if(num / 10 &gt; 0) {
 collectDigits(num / 10, digits);
 }
 digits.add(num % 10);
}
</code></pre>

If you would like to maintain the order of the digits from least significant (index[0]) to most significant (index[n]), the following updated getDigits() is what you need:

<pre><code>/**
 * split an integer into its individual digits
 * NOTE: digits order is maintained - i.e. Least significant digit is at index[0]
 * @param num positive integer
 * @return array of digits
 */
public static Integer[] getDigits(int num) {
 if (num &lt; 0) { return new Integer[0]; }
 List&lt;Integer&gt; digits = new ArrayList&lt;Integer&gt;();
 collectDigits(num, digits);
 Collections.reverse(digits);
 return digits.toArray(new Integer[]{});
}
</code></pre>

blocks|key|968737|text|我认为这将是获取数字的最有用的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|968738|public+int[]+getDigitsOf(int+num)
{++++++++
++++int+digitCount+=+Integer.toString(num).length();

++++if+(num+<+0)+
++++++++digitCount--;+++++++++++

++++int[]+result+=+new+int[digitCount];

++++while+(digitCount--+>0)+{
++++++++result[digitCount]+=+num+%25+10;
++++++++num+/=+10;
++++}++++++++
++++return+result;
}|code-block|syntax|javascript|968739|然后，您可以通过一种简单的方式获取数字：|968740|int+number+=+12345;
int[]+digits+=+getDigitsOf(number);

for+(int+i+=+0;+i+<+digits.length;+i%2B%2B)+{
++++System.out.println(digits[i]);
}|968741|或者更简单地说：|968742|int+number+=+12345;
for+(int+i+=+0;+i+<+getDigitsOf(number).length;+i%2B%2B)+{
++++System.out.println(++getDigitsOf(number)[i]++);
}|968743|注意，最后一个方法调用getDigitsOf方法的时间太长了。所以它会更慢。您应该创建一个整型数组，然后调用一次getDigitsOf方法，就像在第二个代码块中一样。|968744|在下面的代码中，可以反转到process。此代码将所有数字放在一起构成数字：|968745|public+int+digitsToInt(int[]+digits)
{
++++int+digitCount+=+digits.length;
++++int+result+=+0;

++++for+(int+i+=+0;+i+<+digitCount;+i%2B%2B)+{
++++++++result+=+result+*+10;
++++++++result+%2B=+digits[i];
++++}

++++return+result;
}|968746|我提供的两种方法也适用于负数。|968747|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|17|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

I think this will be the most useful way to get digits:

<pre><code>public int[] getDigitsOf(int num)
{ 
 int digitCount = Integer.toString(num).length();

 if (num &lt; 0) 
 digitCount--; 

 int[] result = new int[digitCount];

 while (digitCount-- &gt;0) {
 result[digitCount] = num % 10;
 num /= 10;
 } 
 return result;
}
</code></pre>

Then you can get digits in a simple way:

<pre><code>int number = 12345;
int[] digits = getDigitsOf(number);

for (int i = 0; i &lt; digits.length; i++) {
 System.out.println(digits[i]);
}
</code></pre>

or more simply:

<pre><code>int number = 12345;
for (int i = 0; i &lt; getDigitsOf(number).length; i++) {
 System.out.println( getDigitsOf(number)[i] );
}
</code></pre>

Notice the last method calls getDigitsOf method too much time. So it will be slower. You should create an int array and then call the getDigitsOf method once, just like in second code block.

In the following code, you can reverse to process. This code puts all digits together to make the number:

<pre><code>public int digitsToInt(int[] digits)
{
 int digitCount = digits.length;
 int result = 0;

 for (int i = 0; i &lt; digitCount; i++) {
 result = result * 10;
 result += digits[i];
 }

 return result;
}
</code></pre>

Both methods I have provided works for negative numbers too.

blocks|key|1183197|text|//+could+be+any+num+this+is+a+randomly+generated+one
int+num+=+(int)+(Math.random()+*+1000);

//+this+will+return+each+number+to+a+int+variable
int+num1+=+num+%25+10;
int+num2+=+num+/+10+%25+10;
int+num3+=+num+/100+%25+10;

//+you+could+continue+this+pattern+for+4,5,6+digit+numbers
//+dont+need+to+print+you+could+then+use+the+new+int+values+man+other+ways
System.out.print(num1);
System.out.print("\n"+%2B+num2);
System.out.print("\n"+%2B+num3);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1183198|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>// could be any num this is a randomly generated one
int num = (int) (Math.random() * 1000);

// this will return each number to a int variable
int num1 = num % 10;
int num2 = num / 10 % 10;
int num3 = num /100 % 10;

// you could continue this pattern for 4,5,6 digit numbers
// dont need to print you could then use the new int values man other ways
System.out.print(num1);
System.out.print("\n" + num2);
System.out.print("\n" + num3);
</code></pre>

blocks|key|968866|text|这使用模10方法来计算大于0的数字中的每个数字，然后这将颠倒数组的顺序。这是假设您没有使用"0“作为起始数字。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|968867|这被修改为接受用户输入。这个数组最初是向后插入的，所以我必须使用Collections.reverse()调用将它放回用户的顺序。|offset|length|style|CODE|968868|++++Scanner+scanNumber+=+new+Scanner(System.in);
++++int+userNum+=+scanNumber.nextInt();+//+user's+number

++++//+divides+each+digit+into+its+own+element+within+an+array
++++List<Integer>+checkUserNum+=+new+ArrayList<Integer>();
++++while(userNum+>+0)+{
++++++++checkUserNum.add(userNum+%25+10);
++++++++userNum+/=+10;
++++}

++++Collections.reverse(checkUserNum);+//+reverses+the+order+of+the+array

++++System.out.print(checkUserNum);|code-block|syntax|javascript|968869|entityMap^0|0|W|L|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

This uses the modulo 10 method to figure out each digit in a number greater than 0, then this will reverse the order of the array. This is assuming you are not using "0" as a starting digit.

This is modified to take in user input.
This array is originally inserted backwards, so I had to use the <code>Collections.reverse()</code> call to put it back into the user's order.

<pre><code> Scanner scanNumber = new Scanner(System.in);
 int userNum = scanNumber.nextInt(); // user's number

 // divides each digit into its own element within an array
 List&lt;Integer&gt; checkUserNum = new ArrayList&lt;Integer&gt;();
 while(userNum &gt; 0) {
 checkUserNum.add(userNum % 10);
 userNum /= 10;
 }

 Collections.reverse(checkUserNum); // reverses the order of the array

 System.out.print(checkUserNum);
</code></pre>

blocks|key|1183255|text|我还没有见过任何人使用这种方法，但它对我来说很有效，而且很简短：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1183256|int+num+=+5542;
String+number+=+String.valueOf(num);
for(int+i+=+0;+i+<+number.length();+i%2B%2B)+{
++++int+j+=+Character.digit(number.charAt(i),+10);
++++System.out.println("digit:+"+%2B+j);
}|code-block|syntax|javascript|1183257|这将输出：|1183258|digit:+5
digit:+5
digit:+4
digit:+2|1183259|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I haven't seen anybody use this method, but it worked for me and is short and sweet: 

<pre><code>int num = 5542;
String number = String.valueOf(num);
for(int i = 0; i &lt; number.length(); i++) {
 int j = Character.digit(number.charAt(i), 10);
 System.out.println("digit: " + j);
}
</code></pre>

This will output: 

<pre><code>digit: 5
digit: 5
digit: 4
digit: 2
</code></pre>

blocks|key|1183268|text|int+number+=+12344444;+//+or+it+Could+be+any+valid+number

int+temp+=+0;
int+divider+=+1;

for(int+i+=1;+i<+String.valueOf(number).length();i%2B%2B)
+{

++++divider+=+divider+*+10;

}

while+(divider+>0)+{

++++temp+=+number+/+divider;
++++number+=+number+%25+divider;
++++System.out.print(temp+%2B"+");
++++divider+=+divider/10;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1183269|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>int number = 12344444; // or it Could be any valid number

int temp = 0;
int divider = 1;

for(int i =1; i&lt; String.valueOf(number).length();i++)
 {

 divider = divider * 10;

}

while (divider &gt;0) {

 temp = number / divider;
 number = number % divider;
 System.out.print(temp +" ");
 divider = divider/10;
}
</code></pre>

blocks|key|968955|text|我写了一个程序，演示了如何使用一种更简单、更容易理解的方法来分离整数的数字，这种方法不涉及数组、递归和所有那些花哨的花哨的东西。下面是我的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|968956|int+year+=+sc.nextInt(),+temp+=+year,+count+=+0;

while+(temp>0)
{
++count%2B%2B;
++temp+=+temp+/+10;
}

double+num+=+Math.pow(10,+count-1);
int+i+=+(int)num;

for+(;i>0;i/=10)
{
++System.out.println(year/i%2510);
}|code-block|syntax|javascript|968957|假设您的输入是整数123，结果输出将如下所示：|offset|length|style|CODE|968958|1
2
3|968959|entityMap^0|0|0|9|3|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I wrote a program that demonstrates how to separate the digits of an integer using a more simple and understandable approach that does not involve arrays, recursions, and all that fancy schmancy. Here is my code:

<pre><code>int year = sc.nextInt(), temp = year, count = 0;

while (temp&gt;0)
{
 count++;
 temp = temp / 10;
}

double num = Math.pow(10, count-1);
int i = (int)num;

for (;i&gt;0;i/=10)
{
 System.out.println(year/i%10);
}
</code></pre>

Suppose your input is the integer <code>123</code>, the resulting output will be as follows:

<pre><code>1
2
3
</code></pre>

blocks|key|1183343|text|在这个主题的基础上，下面是如何在Java中确认数字是一个回文整数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1183344|public+static+boolean+isPalindrome(int+input)+{
List<Integer>+intArr+=+new+ArrayList();
int+procInt+=+input;

int+i+=+0;
while(procInt+>+0)+{
++++intArr.add(procInt%2510);
++++procInt+=+procInt/10;
++++i%2B%2B;
}

int+y+=+0;
int+tmp+=+0;
int+count+=+0;
for(int+j:intArr)+{
++++if(j+==+0+&&+count+==+0)+{
++++break;
++++}

++++tmp+=+j+%2B+(tmp*10);
++++count%2B%2B;
}

if(input+!=+tmp)
++++return+false;

return+true;
}|code-block|syntax|javascript|1183345|我确信我可以进一步简化这个算法。然而，这就是我所在的地方它在我的所有测试用例下都能正常工作。|1183346|我希望这对某些人有帮助。|1183347|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Just to build on the subject, here's how to confirm that the number is a palindromic integer in Java:

<pre><code>public static boolean isPalindrome(int input) {
List&lt;Integer&gt; intArr = new ArrayList();
int procInt = input;

int i = 0;
while(procInt &gt; 0) {
 intArr.add(procInt%10);
 procInt = procInt/10;
 i++;
}

int y = 0;
int tmp = 0;
int count = 0;
for(int j:intArr) {
 if(j == 0 &amp;&amp; count == 0) {
 break;
 }

 tmp = j + (tmp*10);
 count++;
}

if(input != tmp)
 return false;

return true;
}
</code></pre>

I'm sure I can simplify this algo further. Yet, this is where I am. And it has worked under all of my test cases.

I hope this helps someone.

blocks|key|1183378|text|import+java.util.Scanner;

public+class+SeparatingDigits+{

++++public+static+void+main(+String[]+args+)
++++{

++++++++System.out.print(+"Enter+the+digit+to+print+separately+:-+");
++++++++Scanner+scan+=+new+Scanner(+System.in+);

++++++++int+element1+=+scan.nextInt();
++++++++int+divider;

++++++++if(+(+element1+>+9999+)+&&+(+element1+<=+99999+)+)
++++++++{
++++++++++++divider+=+10000;
++++++++}
++++++++else+if(+(+element1+>+999+)+&&+(+element1+<=+9999+)+)
++++++++{
++++++++++++divider+=+1000;
++++++++}
++++++++else+if+(+(+element1+>+99)+&&+(+element1+<=+999+)+)
++++++++{
++++++++++++divider+=+100;
++++++++}
++++++++else+if(+(+element1+>+9+)+&&+(+element1+<=+99+)+)
++++++++{
++++++++++++divider+=+10;
++++++++}
++++++++else+
++++++++{
++++++++++++divider+=+1;
++++++++}

++++++++quotientFinder(+element1,+divider+);




++++}

+++++public+static+void+quotientFinder(+int+elementValue,+int+dividerValue+)
+++++{
+++++++++for(+int+count+=+1;++dividerValue+!=+0;+count%2B%2B)
+++++++++{
++++++++++++int+quotientValue+=+elementValue+/+dividerValue+;
++++++++++++elementValue+=+elementValue+%25+dividerValue+;
++++++++++++System.out.printf(+"%25d++",+quotientValue+);

++++++++++++dividerValue+/=+10;

+++++++++}
+++++}
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1183379|而不使用数组和字符串。(数字1-99999+)|unstyled|1183380|输出：|1183381|单独输入要打印的数字:-+12345|1183382|1+2+3+4+5|1183383|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>import java.util.Scanner;

public class SeparatingDigits {

 public static void main( String[] args )
 {

 System.out.print( "Enter the digit to print separately :- ");
 Scanner scan = new Scanner( System.in );

 int element1 = scan.nextInt();
 int divider;

 if( ( element1 &gt; 9999 ) &amp;&amp; ( element1 &lt;= 99999 ) )
 {
 divider = 10000;
 }
 else if( ( element1 &gt; 999 ) &amp;&amp; ( element1 &lt;= 9999 ) )
 {
 divider = 1000;
 }
 else if ( ( element1 &gt; 99) &amp;&amp; ( element1 &lt;= 999 ) )
 {
 divider = 100;
 }
 else if( ( element1 &gt; 9 ) &amp;&amp; ( element1 &lt;= 99 ) )
 {
 divider = 10;
 }
 else 
 {
 divider = 1;
 }

 quotientFinder( element1, divider );




 }

 public static void quotientFinder( int elementValue, int dividerValue )
 {
 for( int count = 1; dividerValue != 0; count++)
 {
 int quotientValue = elementValue / dividerValue ;
 elementValue = elementValue % dividerValue ;
 System.out.printf( "%d ", quotientValue );

 dividerValue /= 10;

 }
 }
 }
</code></pre>

Without using arrays and Strings . ( digits 1-99999 )

output :

Enter the digit to print separately :- 12345

1 2 3 4 5

blocks|key|1183429|text|我认为更简单的方法是将数字转换为字符串，然后使用substring提取，然后转换为整数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1183430|如下所示：|1183431|int+digits1+=Integer.parseInt(+String.valueOf(201432014).substring(0,4));
++++System.out.println("digits+are:+"%2Bdigits1);|code-block|syntax|javascript|1183432|输出为2014年|1183433|entityMap^0|O|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

Easier way I think is to convert the number to string and use <code>substring</code> to extract and then convert to integer.

Something like this:

<pre><code>int digits1 =Integer.parseInt( String.valueOf(201432014).substring(0,4));
 System.out.println("digits are: "+digits1);
</code></pre>

ouput is 
2014

blocks|key|1183460|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1183461|int+num=+4321
int+first++=++num+%25+10;
int+second+=++(+num+-+first+)+%25+100+/+10;
int+third++=++(+num+-+first+-+second+)+%25+1000+/+100;
int+fourth+=++(+num+-+first+-+second+-+third+)+%25+10000+/+1000;|code-block|syntax|javascript|1183462|你将得到first+=+1，second+=+2，third+=3和fourth+=4....|1183463|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|$]]

Try this:

<pre><code>int num= 4321
int first = num % 10;
int second = ( num - first ) % 100 / 10;
int third = ( num - first - second ) % 1000 / 100;
int fourth = ( num - first - second - third ) % 10000 / 1000;
</code></pre>

You will get first = 1, second = 2, third = 3 and fourth = 4 ....

blocks|key|969142|text|public+int[]+getDigitsOfANumber(int+number)+{
++++String+numStr+=+String.valueOf(number);
++++int+retArr[]+=+new+int[numStr.length()];

++++for+(int+i+=+0;+i+<+numStr.length();+i%2B%2B)+{
++++++++char+c+=+numStr.charAt(i);
++++++++int+digit+=+c;
++++++++int+zero+=+(char)+'0';
++++++++retArr[i]+=+digit+-+zero;

++++}
++++return+retArr;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|969143|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 int[] getDigitsOfANumber(int number) {
 String numStr = String.valueOf(number);
 int retArr[] = new int[numStr.length()];

 for (int i = 0; i &lt; numStr.length(); i++) {
 char c = numStr.charAt(i);
 int digit = c;
 int zero = (char) '0';
 retArr[i] = digit - zero;

 }
 return retArr;
}
</code></pre>

blocks|key|969156|text|请看下面我的提案和评论|type|unstyled|depth|inlineStyleRanges|entityRanges|data|969157|++++++++++int+size=i.toString().length();+//+the+length+of+the+integer+(i)+we+need+to+split;
+++++++++++ArrayList<Integer>+li+=+new+ArrayList<Integer>();+//+an+ArrayList+in+whcih+to+store+the+resulting+digits

++++++++Boolean+b=true;+//+control+variable+for+the+loop+in+which+we+will+reatrive+step+by+step+the+digits
++++++++String+number="1";+//+here+we+will+add+the+leading+zero+depending+on+the+size+of+i
++++++++int+temp;++//+the+resulting+digit+will+be+kept+by+this+temp+variable

++++for+(int+j=0;+j<size;+j%2B%2B){
++++++++++++++++++++++++number=number.concat("0");
++++++++++++++++++++}

Integer+multi+=+Integer.valueOf(number);+//+the+variable+used+for+dividing+step+by+step+the+number+we+received+
++++++++++++++++while(b){

++++++++++++++++++++multi=multi/10;
++++++++++++++++++++temp=i/(multi);
++++++++++++++++++++li.add(temp);
++++++++++++++++++++i=i%25(multi);
++++++++++++++++++++++++++++++++++++++++if(i==0){
++++++++++++++++++++++++++++++++++++++++b=false;
++++++++++++++++++++++++++++++++++++++++}


++++++++++++++++}

++++++++++++++++for(Integer+in:+li){
++++++++++++++++++++System.out.print(in.intValue()%2B+"+");
++++++++++++++++}|code-block|syntax|javascript|969158|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

see bellow my proposal with comments

<pre><code> int size=i.toString().length(); // the length of the integer (i) we need to split;
 ArrayList&lt;Integer&gt; li = new ArrayList&lt;Integer&gt;(); // an ArrayList in whcih to store the resulting digits

 Boolean b=true; // control variable for the loop in which we will reatrive step by step the digits
 String number="1"; // here we will add the leading zero depending on the size of i
 int temp; // the resulting digit will be kept by this temp variable

 for (int j=0; j&lt;size; j++){
 number=number.concat("0");
 }

Integer multi = Integer.valueOf(number); // the variable used for dividing step by step the number we received 
 while(b){

 multi=multi/10;
 temp=i/(multi);
 li.add(temp);
 i=i%(multi);
 if(i==0){
 b=false;
 }


 }

 for(Integer in: li){
 System.out.print(in.intValue()+ " ");
 }
</code></pre>

blocks|key|1183563|text|由于我在这个问题上看不到使用Java+8的方法，所以我将把这个放进去。假设您从一个String开始，并希望获得一个List<Integer>，那么您可以像这样流式传输元素。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1183564|List<Integer>+digits+=+digitsInString.chars()
++++++++.map(Character::getNumericValue)
++++++++.boxed()
++++++++.collect(Collectors.toList());|code-block|syntax|javascript|1183565|它以IntStream的形式获取String中的字符，将这些字符的整数表示形式映射到一个数值，对它们进行装箱，然后将它们收集到一个列表中。|1183566|entityMap^0|15|6|1L|D|0|0|2|9|G|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

Since I don't see a method on this question which uses Java 8, I'll throw this in. Assuming that you're starting with a <code>String</code> and want to get a <code>List&lt;Integer&gt;</code>, then you can stream the elements like so.
<pre><code>List&lt;Integer&gt; digits = digitsInString.chars()
 .map(Character::getNumericValue)
 .boxed()
 .collect(Collectors.toList());
</code></pre>
This gets the characters in the <code>String</code> as a <code>IntStream</code>, maps those integer representations of characters to a numeric value, boxes them, and then collects them into a list.

blocks|key|1183641|text|这是我的答案，我是为自己做的，我希望对于那些不想使用字符串方法或需要更数学的解决方案的人来说，它足够简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1183642|public+static+void+reverseNumber2(int+number)+{

++++int+residual=0;
++++residual=number%2510;
++++System.out.println(residual);

++++while+(residual!=number)++{
++++++++++number=(number-residual)/10;
++++++++++residual=number%2510;
++++++++++System.out.println(residual);
++++}
}|code-block|syntax|javascript|1183643|所以我只需要得到单位，打印出来，从数字中减去它们，然后除以10+-它总是没有任何浮动的东西，因为单位没有了，重复。|1183644|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|$]]

Here is my answer, I did it for myself and I hope it's simple enough for those who don't want to use the String approach or need a more math-y solution:

<pre><code>public static void reverseNumber2(int number) {

 int residual=0;
 residual=number%10;
 System.out.println(residual);

 while (residual!=number) {
 number=(number-residual)/10;
 residual=number%10;
 System.out.println(residual);
 }
}
</code></pre>

So I just get the units, print them out, substract them from the number, then divide that number by 10 - which is always without any floating stuff, since units are gone, repeat.

blocks|key|969352|text|从作为字符串的整数中获取int[]形式的数字的Java8解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|969353|int[]+digits+=+intAsString.chars().map(i+->+i+-+'0').toArray();|code-block|syntax|javascript|969354|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Java 8 solution to get digits as int[] from an integer that you have as a String:

<pre><code>int[] digits = intAsString.chars().map(i -&gt; i - '0').toArray();
</code></pre>

blocks|key|969376|text|类似下面这样的代码将返回char[]|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|969377|public+static+char[]+getTheDigits(int+value){
++++String+str+=+"";
++++int+number+=+value;
++++int+digit+=+0;
++++while(number>0){
++++++++digit+=+number%2510;
++++++++str+=+str+%2B+digit;
++++++++System.out.println("Digit:"+%2B+digit);
++++++++number+=+number/10;+++++

++++}
++++return+str.toCharArray();
}|code-block|syntax|javascript|969378|entityMap^0|C|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Something like this will return the <code>char[]</code>:

<pre><code>public static char[] getTheDigits(int value){
 String str = "";
 int number = value;
 int digit = 0;
 while(number&gt;0){
 digit = number%10;
 str = str + digit;
 System.out.println("Digit:" + digit);
 number = number/10; 

 }
 return str.toCharArray();
}
</code></pre>

blocks|key|1183748|text|import+java.util.Scanner;

class++Test+
{++
++++public+static+void+main(String[]+args)+++
++++{++
++++++++Scanner+sc+=+new+Scanner(System.in);+


++++int+num=sc.nextInt();+
++++System.out.println("Enter+a+number+(-1+to+end):"%2Bnum);
++++int+result=0;
++++int+i=0;
++++while(true)+
++++{+
++++++int+n=num%2510;
++++++if(n==-1){
++++++++break;
++++++}
++++++i%2B%2B;
++++++System.out.println("Digit"%2Bi%2B"+=+"%2Bn);
++++++result=result*10%2Bn;
++++++num=num/10;+


++++++if(num==0)+
++++++{+
++++++++break;+
++++++}+
++++}
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1183749|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>import java.util.Scanner;

class Test 
{ 
 public static void main(String[] args) 
 { 
 Scanner sc = new Scanner(System.in); 


 int num=sc.nextInt(); 
 System.out.println("Enter a number (-1 to end):"+num);
 int result=0;
 int i=0;
 while(true) 
 { 
 int n=num%10;
 if(n==-1){
 break;
 }
 i++;
 System.out.println("Digit"+i+" = "+n);
 result=result*10+n;
 num=num/10; 


 if(num==0) 
 { 
 break; 
 } 
 }
 }
}
</code></pre>

blocks|key|969408|text|在Java中，这就是将数字与数字分开并将其存储在数组中的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|969409|public+static+void+main(String[]+args)+{
++++++++System.out.println("Digits+Array::+"%2BArrays.toString(getNumberArr(1100)));
}

private+static+Integer[]+getNumberArr(int+number)+{
++++//will+get+the+total+number+of+digits+in+the+number
++++int+temp+=+number;
++++int+counter+=+0;

++++while+(temp+>+0)+{
++++++++temp+/=+10;
++++++++counter%2B%2B;
++++}
++++//reset+the+temp
++++temp+=+number;

++++//+make+an+array
++++int+modulo;+++++//modulo+is+equivalent+to+single+digit+of+the+number.
++++Integer[]+numberArr+=+new+Integer[counter];
++++for+(int+i+=+counter+-+1;+i+>=+0;+i--)+{
++++++++modulo+=+temp+%25+10;
++++++++numberArr[i]+=+modulo;++
++++++++temp+/=+10;
++++}

++++return+numberArr;
}|code-block|syntax|javascript|969410|输出：|969411|Digits+Array::+[1,+1,+0,+0]|969412|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

in Java, this is how you can separate digits from numbers and store them in an Array.

<pre><code>public static void main(String[] args) {
 System.out.println("Digits Array:: "+Arrays.toString(getNumberArr(1100)));
}

private static Integer[] getNumberArr(int number) {
 //will get the total number of digits in the number
 int temp = number;
 int counter = 0;

 while (temp &gt; 0) {
 temp /= 10;
 counter++;
 }
 //reset the temp
 temp = number;

 // make an array
 int modulo; //modulo is equivalent to single digit of the number.
 Integer[] numberArr = new Integer[counter];
 for (int i = counter - 1; i &gt;= 0; i--) {
 modulo = temp % 10;
 numberArr[i] = modulo; 
 temp /= 10;
 }

 return numberArr;
}
</code></pre>

Output:

<pre><code>Digits Array:: [1, 1, 0, 0]
</code></pre>

blocks|key|1183809|text|我注意到使用Java+8流解决问题的例子很少，但我认为这是最简单的一个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1183810|int[]+intTab+=+String.valueOf(number).chars().map(Character::getNumericValue).toArray();|code-block|syntax|javascript|1183811|需要说明的是:您使用String.valueOf(number)将int转换为字符串，然后使用chars()方法获取一个IntStream+(字符串中的每个字符现在都是一个Ascii数)，然后您需要运行map()方法来获取Ascii数的数值。最后，使用toArray()方法将流更改为int[]数组。|offset|length|style|CODE|1183812|entityMap^0|0|0|A|M|1B|7|2T|5|3I|9|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|N|$]]

I noticed that there are few example of using Java 8 stream to solve your problem but I think that this is the simplest one:

<pre><code>int[] intTab = String.valueOf(number).chars().map(Character::getNumericValue).toArray();
</code></pre>

To be clear:
You use <code>String.valueOf(number)</code> to convert int to String, then <code>chars()</code> method to get an IntStream (each char from your string is now an Ascii number), then you need to run <code>map()</code> method to get a numeric values of the Ascii number. At the end you use <code>toArray()</code> method to change your stream into an int[] array.

blocks|key|1183851|text|作为一个菜鸟，我的答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1183852|String+number+=+String.valueOf(ScannerObjectName.nextInt());+
int[]+digits+=+new+int[number.length()];+
for+(int+i+=+0+;+i+<+number.length()+;+i%2B%2B)
++++int[i]+=+Integer.parseInt(digits.substring(i,i%2B1))|code-block|syntax|javascript|1183853|现在所有的数字都包含在"+digits“数组中。|1183854|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|$]]

As a noob, my answer would be:

<pre><code>String number = String.valueOf(ScannerObjectName.nextInt()); 
int[] digits = new int[number.length()]; 
for (int i = 0 ; i &lt; number.length() ; i++)
 int[i] = Integer.parseInt(digits.substring(i,i+1))
</code></pre>

Now all the digits are contained in the "digits" array.

blocks|key|1183863|text|简单解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1183864|public+static+void+main(String[]+args)+{
++++int+v+=+12345;
++++while+(v+>+0){
++++++++System.out.println(v+%25+10);
++++++++v+/=+10;
++++}
}|code-block|syntax|javascript|1183865|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

simple solution 

<pre><code>public static void main(String[] args) {
 int v = 12345;
 while (v &gt; 0){
 System.out.println(v % 10);
 v /= 10;
 }
}
</code></pre>

blocks|key|969558|text|Java9引入了一种新的Stream.iterate方法，可用于生成流并在特定条件下停止。这可以用来获得数字中的所有数字，使用模数方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|969559|int[]+a+=+IntStream.iterate(123400,+i+->+i+>+0,+i+->+i+/+10).map(i+->+i+%25+10).toArray();|code-block|syntax|javascript|969560|请注意，这将以相反的顺序获取数字，但这可以通过向后循环数组(可悲的是，反转数组是not+that+simple)或通过创建另一个流来解决：|969561|int[]+b+=+IntStream.iterate(a.length+-+1,+i+->+i+>=+0,+i+->+i+-+1).map(i+->+a[i]).toArray();|969562|或|969563|int[]+b+=+IntStream.rangeClosed(1,+a.length).map(i+->+a[a.length+-+i]).toArray();|969564|例如，下面的代码：|969565|int[]+a+=+IntStream.iterate(123400,+i+->+i+>+0,+i+->+i+/+10).map(i+->+i+%25+10).toArray();
int[]+b+=+IntStream.iterate(a.length+-+1,+i+->+i+>=+0,+i+->+i+-+1).map(i+->+a[i]).toArray();
System.out.println(Arrays.toString(a));
System.out.println(Arrays.toString(b));|969566|将打印：|969567|[0,+0,+4,+3,+2,+1]
[1,+2,+3,+4,+0,+0]|969568|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/2137791/1310566^0|C|E|0|0|14|F|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1C|8|@]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|M|3|N|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1H|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1J|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1K|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1L|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1M|8|@]|D|@]|E|$I|J]]|$1|10|3|-4|5|6|7|1N|8|@]|D|@]|E|$]]]|11|$12|$5|13|14|15|E|$16|17]]]]

Java 9 introduced a new <code>Stream.iterate</code> method which can be used to generate a stream and stop at a certain condition. This can be used to get all the digits in the number, using the modulo approach.

<pre><code>int[] a = IntStream.iterate(123400, i -&gt; i &gt; 0, i -&gt; i / 10).map(i -&gt; i % 10).toArray();
</code></pre>

Note that this will get the digits in reverse order, but that can be solved either by looping through the array backwards (sadly reversing an array is <a href="https://stackoverflow.com/a/2137791/1310566">not that simple</a>), or by creating another stream:

<pre><code>int[] b = IntStream.iterate(a.length - 1, i -&gt; i &gt;= 0, i -&gt; i - 1).map(i -&gt; a[i]).toArray();
</code></pre>

or

<pre><code>int[] b = IntStream.rangeClosed(1, a.length).map(i -&gt; a[a.length - i]).toArray();
</code></pre>

As an example, this code:

<pre><code>int[] a = IntStream.iterate(123400, i -&gt; i &gt; 0, i -&gt; i / 10).map(i -&gt; i % 10).toArray();
int[] b = IntStream.iterate(a.length - 1, i -&gt; i &gt;= 0, i -&gt; i - 1).map(i -&gt; a[i]).toArray();
System.out.println(Arrays.toString(a));
System.out.println(Arrays.toString(b));
</code></pre>

Will print:

<pre><code>[0, 0, 4, 3, 2, 1]
[1, 2, 3, 4, 0, 0]
</code></pre>

blocks|key|969598|text|你为什么不这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|969599|String+number+=+String.valueOf(input);
char[]+digits+=+number.toCharArray();|code-block|syntax|javascript|969600|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Why don't you do:
<pre><code>String number = String.valueOf(input);
char[] digits = number.toCharArray();
</code></pre>

blocks|key|1183984|text|试试这个吧。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1183985|const+check+=+(num)+=>+{
++let+temp+=+num
++let+result+=+[]
++while(temp+>+0){
++++let+a+=+temp%2510;
++++result.push(a);
++++temp+=+(temp-a)/10;
++}
++return+result;
}

check(98)+//[+8,+9+]|code-block|syntax|javascript|1183986|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this one.
<pre><code>const check = (num) =&gt; {
 let temp = num
 let result = []
 while(temp &gt; 0){
 let a = temp%10;
 result.push(a);
 temp = (temp-a)/10;
 }
 return result;
}

check(98) //[ 8, 9 ]
</code></pre>

blocks|key|969649|text|一个使用LINQ的.NET解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|969650|List<int>+numbers+=+number.ToString().Select(x+=>+x+-+48).ToList();|code-block|syntax|javascript|969651|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A .NET solution using LINQ.
<pre><code>List&lt;int&gt; numbers = number.ToString().Select(x =&gt; x - 48).ToList();
</code></pre>

I have numbers like 1100, 1002, 1022 etc. I would like to have the individual digits, for example for the first number 1100 I want to have 1, 1, 0, 0.

How can I get it in Java?

How to get the separate digits of an int number?

Java

我有像1100,1002,1022等的数字。我想要单独的数字，例如第一个数字1100，我想有1，1，0，0。我怎样才能在Java中获得它？

问如何获取整型数的独立位数？
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