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Task/Fibonacci-sequence/Apex/fibonacci-sequence.apex

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+/*
+ author: snugsfbay
+ date: March 3, 2016
+ description: Create a list of x numbers in the Fibonacci sequence.
+     - user may specify the length of the list
+     - enforces a minimum of 2 numbers in the sequence because any fewer is not a sequence
+     - enforces a maximum of 47 because further values are too large for integer data type
+     - Fibonacci sequence always starts with 0 and 1 by definition
+*/
+public class FibNumbers{
+
+final static Integer MIN = 2; //minimum length of sequence
+final static Integer MAX = 47; //maximum length of sequence
+
+/*
+  description: method to create a list of numbers in the Fibonacci sequence
+  param: user specified integer representing length of sequence should be 2-47, inclusive.
+      - Sequence starts with 0 and 1 by definition so the minimum length could be as low as 2.
+      - For 48th number in sequence or greater, code would require a Long data type rather than an Integer.
+  return: list of integers in sequence.
+*/
+public static List<Integer> makeSeq(Integer len){
+
+  List<Integer> fib = new List<Integer>{0,1}; // initialize list with first two values
+  Integer i;
+
+  if(len<MIN || len==null || len>MAX) {
+      if (len>MAX){
+          len=MAX; //set length to maximum if user entered too high a value
+      }else{
+          len=MIN; //set length to minimum if user entered too low a value or none
+      }
+  } //This could be refactored using teneray operator, but we want code coverage to be reflected for each condition
+
+  //start with initial list size to find previous two values in the sequence, continue incrementing until list reaches user defined length
+  for(i=fib.size(); i<len; i++){
+    fib.add(fib[i-1]+fib[i-2]); //create new number based on previous numbers and add that to the list
+  }
+
+  return fib;
+  }
+
+}