[LeetCode22]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:

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

Analysis:

recursion, when left parentheses number smaller than n, add it. when right parentheses smaller than left, add it

Java

public class GenerateParentheses {
	ArrayList<String> res;
	StringBuffer pare;
	public List<String> generateParenthesis(int n) {
		res = new ArrayList<>();
		pare = new StringBuffer();
		parenethesis(n, 0, 0, 0);
		return res;
    }
	public void parenethesis(int n, int leftNum, int rightNum, int level){
		if(2*n == level){
			res.add(pare.toString());
			return;
		}
		if(leftNum<n){
			pare.append('(');
			parenethesis(n, leftNum+1, rightNum, level+1);
			pare.deleteCharAt(level);
		}
		if(rightNum<leftNum){
			pare.append(')');
			parenethesis(n, leftNum, rightNum+1, level+1);
			pare.deleteCharAt(level);
		}
	}
}

c++

void CombinationPar(vector<string>& result, string& sample, int deep,   
                           int n, int leftNum, int rightNum)  
  {  
       if(deep == 2*n)  
       {  
           result.push_back(sample);  
           return;         }  
       if(leftNum<n)  
       {  
          sample.push_back('(');  
            CombinationPar(result, sample, deep+1, n, leftNum+1, rightNum);  
            sample.resize(sample.size()-1);  
       }  
       if(rightNum<leftNum)  
       {   
            sample.push_back(')');  
            CombinationPar(result, sample, deep+1, n, leftNum, rightNum+1);  
            sample.resize(sample.size()-1);  
       }  
  }  
  vector<string> generateParenthesis(int n) {  
       // Start typing your C/C++ solution below  
       // DO NOT write int main() function  
       vector<string> result;  
       string sample;  
       if(n!= 0)  
            CombinationPar(result, sample, 0, n, 0, 0);  
       return result;  
  }