Generate Parentheses

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"

像這種類似的題目，通過回溯法來構造是比較好的了。回溯的依據是當前字符是添加左括號，還是添加右括號；爲了能夠正確的添加左右括號，設置curLeft，curRight來分別記錄此時的左、右括號數量，同時要滿足curLeft >= curRight，添加右括號的時候，要注意是否滿足這個條件，結束的條件，應該是不滿足這個條件，或者curRight == n構造一個完畢。

class Solution {
public:
    //回溯法
    void generateP(int n, int curLeft, int curRight, string curStr, vector<string> &ans)
    {
        if(curLeft < curRight)
            return;
        if(curRight == n)
        {
            ans.push_back(curStr);
            return;
        }
        
        //添加左邊括號
        if(curLeft < n)
        {
            curStr.push_back('(');
            generateP(n, curLeft + 1, curRight, curStr, ans);
            curStr.pop_back();
        }
        
        //添加右邊括號
        if(curRight < n && curRight < curLeft)
        {
            curStr.push_back(')');
            generateP(n, curLeft, curRight + 1, curStr, ans);
            curStr.pop_back();
        }
    }
    
    vector<string> generateParenthesis(int n) {
        vector<string> ans;
        if(n > 0)
        {
            string curStr = "";
            generateP(n, 0, 0, curStr, ans);
        }
        
        return ans;
    }
};