### Problem Statement

```Class Name: FriendlyIntParser
Method Name: parseInt
Parameters: String
Returns: int

Computers have made their way into a significant percentage of the homes in
America, and as programmers we are responsible for designing user interfaces
that everyone can use.  User interfaces need to be flexible so that if a user
makes some non-fatal error, the interface can still figure out what the user
meant.

Your task is to write a class FriendlyIntParser which processes text input from
a webpage.  The input should be a String representation of an integer, however,
because this is a friendly user interface, we cut the user a little more slack:

1. If the user types the letter "O" or "o", we assume that they meant the digit
"0".
2. If the user types the letter "l", we assume that they meant the digit "1".
3. Commas and spaces are allowed, but are not processed (ignore them).

If the user has still not entered a valid positive integer (even with the rules
above), return -1 indicating an error.  Overflow is considered invalid, and -1
should be returned.  Note:  A value greater than 2147483647 is overflow in both
C++ and Java.

Here is the method signature (be sure your method is public):
int parseInt(String n);

Input: n will contain between 0 and 50, inclusive, letters, numbers, spaces, or
commas.

Returns:  the integer represented by n, or -1 if n is not a valid non-negative
integer

Examples:
"lo6" returns 106.
"234,657" returns 234657.
"hi" returns -1.
",,,,,5,,5,  4" returns 554.
"2200000000" returns -1.
```

### Definition

 Class: FriendlyIntParser Method: parseInt Parameters: String Returns: int Method signature: int parseInt(String param0) (be sure your method is public)

#### Problem url:

http://www.topcoder.com/stat?c=problem_statement&pm=144

#### Problem stats url:

http://www.topcoder.com/tc?module=ProblemDetail&rd=3023&pm=144

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