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JavaScript print star pyramid function

小云云Original: 2018-01-30 13:19:473419browse

This article mainly introduces JavaScript to realize the function of printing star pyramid. It combines specific examples to analyze the principle and related implementation techniques of JavaScript for outputting any given number of rows of star pyramid graphics. Friends who need it can refer to it. I hope it can help. Everyone.

It is exactly the same for you to write in other languages.

I guess everyone has seen this question when they were learning C language...

That is, print the following The ghost thing:

When I saw the loop structure, I felt very bored, so I disdained this question and didn’t think about it carefully, because

If it is written in JavaScript, it can be written like this. It is not even JavaScript, it is just an html:

<html>
<head>
<meta http-equiv="content-type" content="text/html;charset=gb2312"/>
</head>
<body>
＆nbsp;＆nbsp;*
＆nbsp;***
*****
＆nbsp;***
＆nbsp;＆nbsp;*
</body>
</html>

The reason why utf-8 encoding is not used here is because utf-8 is suitable for nbsp's processing and * font will cause layout confusion, that is, the standard Song font is not used, making the final result impossible to view.

Anyway, I can copy and paste the above content no matter how many lines you put in the question. However, today, when I have programmed to a certain amount and have some big data concepts, I saw this question again and carefully After thinking about it for a while, if the question maker asked to output a star-shaped pyramid with 200,000 lines of symmetry on the central axis, I would be damned.

So we still need to thoroughly understand how to implement this. Although such programming will not occur in actual programming, it is said that this question will still come up in some boring interviews. At least, latecomers will ask you C language When asked, if you output a star pyramid with 200,000 lines of symmetry on the central axis, you still have to know how to do it. If you can't do this question the first time, you can't do it.

1. Basic Goals

First, an input box pops up, allowing the user to enter an odd number. After all, axial symmetry requires an odd number,
Then for the sake of the robustness of the program, it is necessary to judge what the user inputs. If the input is not an odd number, a prompt will pop up and the subsequent program will not be executed. How to judge an odd number in JavaScript? I have already discussed it in "JavaScript's Judgment and Processing of Numbers" ” has been said before, so I won’t go into details here.

Considering the load of the browser, here, I only allow the odd number entered by the user to be 189. You can also increase it a bit. 189 feels okay in my computer, so I set this number. It has no special meaning. It was entered randomly and was not intentionally tested.

Enter a line of 189, and IE has popped up a prompt to "abort the script", but there is no problem if you don't abort!

The running results are as follows:

If you are writing a C language program or other programs, this value can definitely be set larger!

2. Basic Idea

Once you understand this, it is very easy to write.

First of all, we only need to output spaces on the left side of *. There is no need to output spaces on the right side. After typing *, just change the line directly.

Split into two Parts, one part is when i739fcc98bde448d16560a3bf646b3798n/2 in the lower half.

The reason why it is divided like this is because the * sign in the output of these two parts It is different from the number of spaces output.

After that, it was a common problem in junior high school to find the rules. Anyway, I found the above rules, and there was no problem when programming. Other math emperors found more awesome expressions, and I was defeated.

3. Production process

#The code is very simple, it is the expression of conditional structure and loop structure. It goes without saying that the above ideas are understood. .

Some people may find it strange here, why do I need to judge n++ first and then n%2!=0, that is, judging whether n+1 is an even number to judge whether n is an odd number,

The main thing here is It is to cater to the following for loop structure...

<html>
<head>
<meta http-equiv="content-type" content="text/html;charset=gb2312"/>
</head>
<body>
</body>
</html>
<script>
var i,j,k,n;
n=window.prompt("请输入要输出的行数n，为了形成轴对称，所以你输出的必须是奇数！");
if(isNaN(n)||!n)
  alert("你输入的不是数！");
else{
  n++;
  if(n%2!=0)
    alert("你输入的不是奇数！");
  else if(n>190){
    alert("不要这么大嘛！臣妾做不到啊！");
  }
  else{
    for(i=1;i<n;i++){
      if(i<=n/2){
        for(k=n/2-i;k>0;k--)
          document.write("＆nbsp;");
        for(j=0;j<2*i-1;j++)
          document.write("*");
          }
      else{
        for(k=i-n/2;k>0;k--)
          document.write("＆nbsp;");
        for(j=0;j<2*(n-i)-1;j++)
          document.write("*");
        }
      document.write("<br>");
    }
  }
}
</script>

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