reverse an integer (in a crazy way)

<br />public static int reverse(int i) {<br /> i = (i & 0x55555555) <<>>> 1) & 0x55555555;<br /> i = (i & 0x33333333) <<>>> 2) & 0x33333333;<br /> i = (i & 0x0f0f0f0f) <<>>> 4) & 0x0f0f0f0f;<br /> i = (i <<>>> 8) & 0xff00) | (i >>> 24);<br /> return i;<br />}<br />


compare this with my really old post about the same problem

Java: find nested balanced tags

Problem:
I needed to parse out some divs from a html page, and java's regex cant do it. atleast when you have arbitrary number of nested balanced elements.

Solution:
<br /><br />private String removeDivContents(String id, String clazz,<br /> String text){<br />String div = "<div class="\"><br /> +clazz.toLowerCase()+"\" id=\""+id.toLowerCase()+"\">";<br />String textl = text.toLowerCase();<br />int index = textl.indexOf(div);<br />if(index == -1){<br /> return text;<br />}<br />int loc = index + div.length();<br />int end = findBalancedClosingTag(loc, textl,<br /> "<div","</div>", 1);<br />return text.substring(0,index) + text.substring(end);<br />}<br /><br />private int findBalancedClosingTag(int start,<br />String textl, String startTag, String endTag, int unClosedStarts){<br />int index = textl.indexOf(endTag,start);<br />if(index == -1){<br /> return start;<br />}<br />Pattern p = Pattern.compile(startTag);<br />Matcher m = p.matcher(textl.substring(start, index));<br />unClosedStarts --; // we are in this function so we closed ONE start tag<br />while(m.find()){<br /> unClosedStarts ++;<br />}<br />if(unClosedStarts == 0){<br /> return index + endTag.length();<br />} else {<br /> return findBalancedClosingTag(index + endTag.length(), textl, startTag, endTag, unClosedStarts);<br />}<br />}<br />
Use at your own risk.

quick mod by 11

1==1 (mod 11)
10==-1 (mod 11)
100==1 (mod 11)
1000==-1 (mod 11)

Therefore 123456 == 6 - 5 + 4 - 3 + 2 - 1 (mod 11) == 3 (mod 11)
Obviously you may need to do a further % operation after adding and subtracting digits, but it should be reasonably fast.

A cool function

I wish i saw this when i had to do java homework, it would have helped.
to bad i never had to since reverse any numbers.
<br />int getReverse(int i) {<br /> int j = 0;<br /> while (i != 0) {<br /> j *= 10;<br /> j += i % 10;<br /> i /= 10;<br /> }<br /> return j;<br />}<br />