Project Euler Problem 14: Longest Collatz sequence

Problem

The following iterative sequence is defined for the set of positive integers:

n → n/2 (n is even)
n → 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

Solution

There are two tricks

• Use long instead of int
• Save the result for each number along the way to an array

      private static final int NUMBER = 1000000;  
      private static final int[] TERMS = new int[NUMBER];  
      public static void main(String[] args) {  
           long largest = 0;  
           int num = 0;  
           for (int i=1; i<NUMBER; i++){  
                long terms =terms(i);   
                if (terms > largest){  
                     largest = terms;  
                     num = i;  
                }  
           }  
           System.out.println(num+" "+largest);  
      }  
      private static long next(long n){  
           if (n % 2 == 0){  
                return n / 2;  
           }else{  
                return 3*n+1;  
           }  
      }  
      private static int terms(long n){  
           long temp = n;  
           int count = 1;  
           while (temp > 1){  
                temp = next(temp);  
                if (temp < NUMBER && TERMS[(int)temp] != 0){  
                     count += TERMS[(int)temp];  
                     break;  
                }  
                count++;  
           }  
           TERMS[(int)n] = count;  
           return count;  
      }