3206. Alternating Groups I

Solution

class Solution {
public:
    int numberOfAlternatingGroups(vector<int>& colors)
    {
        int n = colors.size();
        for (int i = 0; i < n; ++i)
        {
            colors.push_back(colors[i]);
        }

        int cnt = 0;
        for (int i = 0; i < n; ++i)
        {
            if (colors[i] == colors[i + 2] && colors[i] != colors[i + 1])
            {
                ++cnt;
            }
        }
        return cnt;
        
    }
};

• 時間複雜度：
  $O(n)$
• 空間複雜度：
  $O(1)$