Cyclomatic complexity - draw control flow graph for this java statement

Can anyone help with this?

while (x > level)
    x = x – 1;
x = 0

Cyclomatic complexity can be computed using the formula provided here.

Cyclomatic complexity = E - N + P 
where,
  E = number of edges in the flow graph.
  N = number of nodes in the flow graph.
  P = number of nodes that have exit points

For your case, the graph should look like this:

---------------                ----------
|  x > level  |----- NO ------>| x = x-1|
|-------------|                ----|-----
       |      |---------------------
       |
      Yes
       |
-------|----------
| End while (if) |
-------|----------
       |
       |
   ---------
   |  x = 0 |
   ----------

(not an ASCII art person)

So, the cyclomatic complexity should be:

E = 4, N = 4, P = 2 => Complexity = 4 - 4 + 2 = 2

[edit] Ira Baxter points out very well how to simplify this computation for languages like Java, C#, C++ etc. However, identifying the conditionals must be carefully performed, as shown here:

- Start with a count of one for the method.
- Add one for each of the following flow-related elements that are found in the method.
    Returns - Each return that isn't the last statement of a method.
    Selection - if, else, case, default.
    Loops - for, while, do-while, break, and continue.
    Operators - &&, ||, ?, and :
    Exceptions - catch, finally, throw, or throws clause.