Bouncy Numbers

Recently I have done a lot of Project Euler problems and I solved problem 112.

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.

Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.
Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.

Find the least number for which the proportion of bouncy numbers is exactly 99%.

 I found that there are two ways to solve this problem efficiently.  The first takes two passes through the number to determine if it is bouncy.  The other takes only one pass.  The first method determines if the number is increasing, then decreasing, and if it is neither then it is bouncy.  The second just determines if it is bouncy.

To determine if a number is increasing or decreasing, I used the same algorithm but just switched the comparison operator. The Algorithm is starting from the right hand side of the number compare the last two digits, for increasing the digit prior should be less than the succeeding digit and vice versa for decreasing.
Psuedo-Code:
  
  i //i is the number we are checking
  
  while(i > 9){
    j = i mod 10
    i /= 10
    k = i mod 10
    if(k < j)  //This checks if the number is decreasing.
      return false
  }
  
  return true


The other method uses one pass to check if the number is bouncy.
It checks the difference between the digits and checks for a switch in the sign.
If the number is increasing (12345) the difference between the digits, subtracting the right from the left (2 - 3 = -1), is always negative. Decreasing the difference is always positive.  If the number is bouncy then there will be a change a of sign.  Detecting that change of sign can be done with an xor.  The only tricky part is dealing with zeros.  Zeros have to be skipped over because they can be apart of both increasing and decreasing numbers.

Psuedo-Code:

  

  i

  r = 0




  while(r == 0){ // gets r to first non-zero

    j = i mod 10

    i /= 10
    k = i mod 10
    r = k - j
  while(i > 9){
    j = i mod 10
    i /= 10    k = i mod 10
    p = k - j
    if(p xor r < 0 )  //This checks for a sign change
      return true
  }

  return false

These are both fast but the second method is 60% faster than the first method.