blocks|key|718134|text|我们可以观察到数字1只打印一次，2打印两次，3打印三次，等等。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|718135|firstValueInLine保持行的开始编号；number是打印的行中的编号；counter只确保number打印number次|718136|++++for+(int+firstValueInLine+=+1;+firstValueInLine+<=+5;+%2B%2BfirstValueInLine)+{
++++++++for+(int+number+=+firstValueInLine;+number+<=+5;+%2B%2Bnumber)+{
++++++++++++for+(int+counter+=+0;+counter+<+number;+%2B%2Bcounter)+{
++++++++++++++++System.out.print(number);
++++++++++++}
++++++++}
++++++++System.out.println();
++++}|code-block|syntax|javascript|718137|entityMap^0|9|1|G|1|M|1|0|0|G|P|6|15|7|1F|6|1N|6|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|N|$]]

We can use the observation that number <code>1</code> is printed only once, <code>2</code> twice, <code>3</code> thrice, etc.

<code>firstValueInLine</code> keeps the number the line starts;
<code>number</code> is the number in line being printed;
<code>counter</code> just ensures that <code>number</code> is printed <code>number</code> times

<pre><code> for (int firstValueInLine = 1; firstValueInLine &lt;= 5; ++firstValueInLine) {
 for (int number = firstValueInLine; number &lt;= 5; ++number) {
 for (int counter = 0; counter &lt; number; ++counter) {
 System.out.print(number);
 }
 }
 System.out.println();
 }
</code></pre>

blocks|key|954229|text|您在行的开头以及一个数字(这里是字符)的重复次数上犯了错误。通过以下方式进行修复：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|954230|for(int+i+=+1;+i+<=+5;+i%2B%2B)+{++++++++++//+Each+iteration+for+one+line
++++for(int+a+=+i;+a+<=+5;+a%2B%2B)+{++++++//+starts+with+a+for+ith+line
++++++++for(int+b+=+1;+b+<=+a;+b%2B%2B)+{++//+a+times+`a`+digit
++++++++++++System.out.print(a);
++++++++}
++++}
++++System.out.println();
}|code-block|syntax|javascript|954231|954232|要简单地你的问题，首先考虑打印这个模式：|954233|12345
2345
345
45
5|954234|然后对其进行扩展:在最内层的循环中，将重复代码放入等于数字次数的位置，使用：|offset|length|style|BOLD|954235|for(int+b+=+1;+b+<=+a;+b%2B%2B)+{++//+a+times+`a`+digit
+++++System.out.print(a);
}|954236|entityMap^0|0|0|0|0|0|2|G|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|K|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|L|3|M|5|6|7|10|8|@$N|11|O|12|P|Q]]|9|@]|A|$]]|$1|R|3|S|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|T|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|U|$]]

You made mistake in how the line starts and how many times a digit (here character) gets repeated. Fix it by:

<pre><code>for(int i = 1; i &lt;= 5; i++) { // Each iteration for one line
 for(int a = i; a &lt;= 5; a++) { // starts with a for ith line
 for(int b = 1; b &lt;= a; b++) { // a times `a` digit
 System.out.print(a);
 }
 }
 System.out.println();
}
</code></pre>

<hr>

To simply your problem, first think about printing this pattern :

<pre><code>12345
2345
345
45
5
</code></pre>

Then extend it: in innermost loop put the code for repetition equal to digit times, using :

<pre><code>for(int b = 1; b &lt;= a; b++) { // a times `a` digit
 System.out.print(a);
}
</code></pre>

blocks|key|954262|text|这里的事情是，使用第一个for-loop来表示行数，使用第二个for-loop来获取要打印的数字。3for-loop打印所需数量的显示数字。相反，你所做的事情是正确的。所以让你的代码颠倒过来。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|954263|尝尝这个|954264|public+static+void+problemFour()+{
++++++for(int+i+=+1;+i+<=+5;+i%2B%2B)+{
+++++++++for(int+a+=+i;+a+<=+5;+a%2B%2B)+{
+++++++++++++for(int+b=0;b<a;b%2B%2B){
+++++++++++++++System.out.print(a);++
+++++++++++++}++++++++++++++++
+++++++++}
+++++++++System.out.println();
++++++}
+++}|code-block|syntax|javascript|954265|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Thing here is, use 1st for-loop for number of lines, 2nd for-loop for getting the number to print. 3rd for-loop to print the displaying number for the required amount.
What you've done is correct in opposite way. So make your code reverse. 

Try this

<pre><code>public static void problemFour() {
 for(int i = 1; i &lt;= 5; i++) {
 for(int a = i; a &lt;= 5; a++) {
 for(int b=0;b&lt;a;b++){
 System.out.print(a); 
 } 
 }
 System.out.println();
 }
 }
</code></pre>

blocks|key|954267|text|查看内联注释。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|954268|public+static+void+main(String[]+args)+{
++++for+(int+i+=+5,+j+=+1;+i+>=+1;+i--,+j%2B%2B)+{+//+Introduce+a+new+variable+j+
++++++++for+(int+a+=+j;+a+<=+5;+a%2B%2B)+{+//+change+a=1+to+a=j+&+a<=i+to+a<=5
++++++++++++for+(int+b+=+1;+b+<=+a;+b%2B%2B)+{
++++++++++++++++System.out.print(a);
++++++++++++}
++++++++}
++++++++System.out.println();
++++}
}|code-block|syntax|javascript|954269|输出：|954270|122333444455555
22333444455555
333444455555
444455555
55555|954271|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|$]]

Checkout the inline comments.

<pre><code>public static void main(String[] args) {
 for (int i = 5, j = 1; i &gt;= 1; i--, j++) { // Introduce a new variable j 
 for (int a = j; a &lt;= 5; a++) { // change a=1 to a=j &amp; a&lt;=i to a&lt;=5
 for (int b = 1; b &lt;= a; b++) {
 System.out.print(a);
 }
 }
 System.out.println();
 }
}
</code></pre>

Output: 

<pre><code>122333444455555
22333444455555
333444455555
444455555
55555
</code></pre>

blocks|key|954273|text|您可以尝试以下操作。看起来不错。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|954274|++public+static+void+problemFour()+{
++++for+(int+i+=+1;+i+<=+5;+i%2B%2B)+{
++++++for+(int+a+=+i;+a+<=+5;+a%2B%2B)+{
++++++++for+(int+b+=+1;+b+<=+a;+b%2B%2B)+{
++++++++++System.out.print(a);
++++++++}
++++++}
++++++System.out.println();
++++}
++}|code-block|syntax|javascript|954275|输出：|offset|length|style|BOLD|954276|122333444455555
22333444455555
333444455555
444455555
55555|954277|entityMap^0|0|0|0|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|$]]

You can try the following. It will work.

<pre><code> public static void problemFour() {
 for (int i = 1; i &lt;= 5; i++) {
 for (int a = i; a &lt;= 5; a++) {
 for (int b = 1; b &lt;= a; b++) {
 System.out.print(a);
 }
 }
 System.out.println();
 }
 }
</code></pre>

Output:

<pre><code>122333444455555
22333444455555
333444455555
444455555
55555
</code></pre>

I need to write some java using 3 "for" loops that outputs

122333444455555

22333444455555

333444455555

444455555

55555

The code I have so far:

<pre><code>public static void problemFour() {
 for(int i = 5; i &gt;= 1; i--) {
 for(int a = 1; a &lt;= i; a++) {
 for(int b = 1; b &lt;= a; b++) {
 System.out.print(a);
 }
 }
 System.out.println();
 }
 }
</code></pre>

This outputs

<pre><code>111112222333445
11111222233344
111112222333
111112222
11111
</code></pre>

I've switched around a lot of combinations of ++'s --'s, &lt;'s, >'s, 5's, and 1's.

I'm pretty stuck, if someone could point me in the right direction, that would be fantastic.

Triple Nested For Loop with specific output (java)

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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我需要使用3个"for“循环来编写一些java输出1223334444555552233344445555533344445555544445555555555到目前为止，我的代码如下：public static void problemFour() {      for(int i = 5; i >= 1; i--)...

问具有特定输出的三重嵌套For循环(java)
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问具有特定输出的三重嵌套For循环(java)EN