Wiggle Subsequence Solutions in C++

// Source : https://leetcode.com/problems/wiggle-subsequence/
// Author : Calinescu Valentin
// Date   : 2016-08-08


 
 /* Solution
  * --------
  * 1) O(N)
  * 
  * We notice that adding a new number to an existing subsequence means finding one that
  * is smaller or bigger than the previous number, according to the difference between the
  * previous number and the number before that as we always need to alternate between increasing
  * and decreasing subsequences. If we encounter increasing or decreasing sequences of 2 or 
  * more consecutive numbers we can treat the entire subsequence as a number, because that way 
  * we can always be sure we don't miss any solution, as finding a number smaller than any 
  * number of an increasing subsequence is guaranteed to be smaller than the biggest number 
  * in the subsequence. Thus, we can only check the difference between consecutive numbers.
  * 
  * Follow up:
  * 
  * The time complexity is already O(N).
  */
class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int solution = 0;//if we have an empty vector the solution is 0
        if(nums.size())
        {
            solution = 1;
            int bigger = 0;//0 is the starting point to be followed by either an increasing or decreasing sequence
            for(int i = 1; i < nums.size(); i++)
            {
                if(nums[i] == nums[i - 1])
                    continue;//we can ignore duplicates as they can always be omitted
                else if(nums[i] > nums[i - 1])
                {
                    if(bigger == 0 || bigger == 2)
                    {
                        bigger = 1;//1 means we now have an increasing sequence
                        solution++;
                    }
                }
                else //if(nums[i] < nums[i - 1])
                {
                    if(bigger == 0 || bigger == 1)
                    {
                        bigger = 2;//2 means we now have a decreasing sequence
                        solution++;
                    }
                }
            }
        }
        return solution;
    }
};