22. 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:

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

【思路】设置两个计数器，分别计数左右括号的数量，左括号初始为n，表示需要产生ｎ个左括号，右括号为０。如果当前选择左括号，则left计数减一，并且右括号数量加１，表示当前需要产生right个右括号来匹配左括号。当左右计数器都为０，则表明已经把左右括号全部产生完，此分支结束。

class Solution {
public:
    void DFS(vector<string> &result, string str, int left, int right)
    {
        if(left==0 && right ==0){
            result.push_back(str);
            return;
        }
        if(left > 0)
        {
            DFS(result, str+"(", left-1, right +1 );
        }
        if(right > 0)
        {
            DFS(result, str+")", left, right-1);
        }
    }
    vector<string> generateParenthesis(int n) {
        vector<string> result;
        if(n<=0) return result;
        string str;
        DFS(result, str, n,0);
        return result;
    }
};