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/**
 * 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:
 * "((()))", "(()())", "(())()", "()(())", "()()()"
 */
 public class Solution {
    public ArrayList<String> generateParenthesis(int n) {
        if(n == 0)
            return new ArrayList<String>();
        return generateParenthesis(n, 0, 0, new StringBuffer());
    }
    
    // overloading...
    public ArrayList<String> generateParenthesis(int n, int l, int r, StringBuffer buf){
        ArrayList<String> res = new ArrayList<String>();
        if(l > n || r > n)
            return res;
        if(r == n){
            res.add(buf.toString());
            return res;
        }
        // adding open parenthesis
        if(l < n){
            StringBuffer newBuf = new StringBuffer(buf);
            newBuf.append("(");
            res.addAll(generateParenthesis(n, l + 1, r, newBuf));
        }
        // adding close parenthesis
        if(r < l){
            StringBuffer newBuf = new StringBuffer(buf);
            newBuf.append(")");
            res.addAll(generateParenthesis(n, l, r + 1, newBuf));
        }
        return res;
    }
    
    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<String>();
        String item = new String();
        if (n <= 0) {
            return res;
        }
        generate(res, item, n, n);
        return res;
    }
    
    public void generate(List<String> res, String item, int left, int right) {
        if (left > right) {
            return;
        }
        if (left == 0 && right == 0) {
            res.add(new String(item));
        }
        if (left > 0) {
            generate(res, item + "(", left - 1, right);
        } 
        if (right > 0) {
            generate(res, item + ")", left, right - 1);
        }
    }
 }