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:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

 

【Leet Code】22. Generate Parentheses_字符串

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

【Leet Code】22. Generate Parentheses_LeetCode_02

传字符串,变成传字符串指针,节约了一半时间
class Solution {
public:
		vector<string> generateParenthesis(int n) {
			vector<string> result;
			string temp = "";
			dfsGP(result, temp, n, n, n);
			return result;
		}
		void dfsGP(vector<string>& result, string& cur, int n, int lN, int rN) {
			if (!lN) {
				for (int i = 0; i < rN; ++i)
					cur.push_back(')');
				result.push_back(cur);
				for (int i = 0; i < rN; ++i)
                    cur.pop_back();
			}
			else {
                cur.push_back('(');
				dfsGP(result, cur, n, lN - 1, rN);
                cur.pop_back();
				if (rN > lN) {
                    cur.push_back(')');
					dfsGP(result, cur, n, lN, rN - 1);
                    cur.pop_back();
				}
			}
		}
};

【Leet Code】22. Generate Parentheses_回溯算法_03