Java 将整数转换为等效数量的空格
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Convert integer to equivalent number of blank spaces
提问by mike
I was wondering what the easiest way is to convert an integer to the equivalent number of blank spaces. I need it for the spaces between nodes when printing a binary search tree. I tried this
我想知道最简单的方法是将整数转换为等效数量的空格。打印二叉搜索树时，我需要它作为节点之间的空间。我试过这个
int position = printNode.getPosition();
String formatter = "%1"+position+"s%2$s\n";
System.out.format(formatter, "", node.element);
But I am getting almost 3 times as many spaces compared to the int value of position. I'm not really sure if I am formatting the string right either. Any suggestions would be great! If it makes it clearer, say position = 6; I want 6 blank spaces printed before my node element.
但是与位置的 int 值相比，我得到的空间几乎是原来的 3 倍。我也不确定我是否正确格式化了字符串。任何建议都会很棒！如果它更清楚，请说 position = 6; 我想在我的节点元素之前打印 6 个空格。
采纳答案by Eyal Schneider
I think you meant something like:
我想你的意思是这样的：
int n = 6;
String s = String.format("%1$"+n+"s", "");
System.out.format("[%13s]%n", ""); // prints "[ ]" (13 spaces)
System.out.format("[%1s]%n", ""); // prints "[ ]" (3 spaces)
回答by Oak
Why not loop over the integer and add a space on each iteration?
为什么不循环遍历整数并在每次迭代时添加一个空格？
String spaces = "";
for (int i = 0 ; i < position ; i++) spaces += " ";
And you can use StringBuilderinstead of String, if positionmight get very big and performance is an issue.
StringBuilder如果position可能会变得非常大并且性能是一个问题，您可以使用代替字符串。
回答by polygenelubricants
You can create a char[]of the desired length, Arrays.fillit with spaces, and then create a Stringout of it (or just appendto your own StringBuilderetc).
您可以创建char[]所需长度的 a ，Arrays.fill它带有空格，然后从中创建一个String（或仅创建append您自己的StringBuilder等）。
import java.util.Arrays;
int n = 6;
char[] spaces = new char[n];
Arrays.fill(spaces, ' ');
System.out.println(new String(spaces) + "!");
// prints " !"
If you're doing this with a lot of possible values of n, instead of creating and filling new char[n]every time, you can create just one long enough string of spaces and take shorter substringas needed.
如果您使用 的许多可能值来执行此操作n，则无需new char[n]每次都创建和填充，您可以只创建一个足够长的空格字符串并substring根据需要缩短。
回答by Sylar
Straight foward solution:
直接解决方案：
int count = 20;
StringBuilder sb = new StringBuilder(count);
for (int i=0; i < count; i++){
sb.append(" ");
}
String s = sb.toString();
StringBuilder is efficient in terms of speed.
StringBuilder 在速度方面是高效的。
回答by Martijn Courteaux
This is an easy, but rubbish, way:
这是一种简单但很垃圾的方法：
int count = 20;
String spaces = String.format("%"+count+"s", "");
or filled in
或填写
String spaces = String.format("%20s", "");
回答by Gui
Here's an example:
下面是一个例子：
public class NestedLoop {
public static void main (String [] args) {
int userNum = 0;
int i = 0;
int j = 0;
while (i<=userNum) {
System.out.println(i);
++i;
if (i<=userNum) {
for (j=0;j<i;j++) {
System.out.print(" ");
}
}
}
return;
}
}
If we make userNum = 3, the output would be:
如果我们让 userNum = 3，输出将是：
0
1
2
3
Hope it would help!
希望它会有所帮助！
回答by Bill K
If one were going to use an iterative solution anyway, one might want to do it a more "Groovy" way, like:
如果无论如何都打算使用迭代解决方案，那么人们可能希望以一种更“Groovy”的方式来实现，例如：
def spaces=6
print ((1..spaces).collect(" ").join())
Note that this doesn't work for zero spaces. for that you might want something more like:
请注意，这不适用于零空格。为此，您可能想要更像：
print (spaces?(1..spaces).collect(" ").join():"")
回答by Subu
//This prints the spaces according to the number in the variable n. So in this case it will print 15 spaces.
//这将根据变量n中的数字打印空格。所以在这种情况下，它将打印 15 个空格。
int n=15;
整数 n=15;
System.out.format("%1$"+n+"s", "");
System.out.format("%1$"+n+"s", "");
