22. Generate Parentheses.txt

//Ìîé êîä
//Ïðîñòàÿ ðåêóðñèÿ. Îñíîâíàÿ ôóíêöèÿ âûçûâàåò ôóíêöèþ ðåêóðñèè. Â îñíîâíîé ôóíêöèè îïèñàíà ñòðîêà â êîòîðîé íàõîäÿòñÿ ñêîáêè "()".
// êàæäîé ôóíêöèè ðåêóðñèè ñíà÷àëà ïðîâåðÿåì óñëîâèÿ âûïîëíåíèÿ. Îòñåèâàåì íåâåðíûå âûâîäû. Åñëè âñå øàãè ïðîéäåíû òî çàïèñûâàåì îòâåò â âåêòîð.
//Äëÿ ðàçâåòâëåíèÿ â êàæäîé ôóíêöèè åñòü öèêë ïðîõîäÿùèé ïî îáîèì ñêîáêàì.
class Solution {
public:
    void recursion(int n, int left,int right, int stepsNum, string variation, 
        vector<string>& answer, string parentheses) {
        if (stepsNum == 0  && variation[variation.size() - 1] != '(')
                answer.push_back(variation);
        if (left <= n && left >= right && stepsNum > 0)
        {
            for(char parenthesis : parentheses)
            {
                if (parenthesis == '(')
                    recursion(n, left + 1, right, stepsNum - 1, variation + parenthesis, answer, parentheses);
                else
                    recursion(n, left, right + 1, stepsNum - 1, variation + parenthesis, answer, parentheses);
            }
        }
    }
    vector<string> generateParenthesis(int n) {
        string parentheses = { "()" };
        int left = 1, right = 0;
        int stepsNum = n * 2 - 1;
        string variation = "(";
        vector<string> answer;
        recursion(n, left, right, stepsNum, variation, answer, parentheses);
        return answer;
    }
};