leetcode--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:

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

public class Solution {
    /**In order to generate valid parenthesis, the number of 
	 * "(" should not smaller than the number of ")". So, we use a List<Integer>
	 * to save the number of "(". So the number of ")" can be calculated.
	 * @param n --Integer, number of parentheses
	 * @return List of valid parentheses strings
	 * @author Averill Zheng
	 * @version 2014-06-04
	 * @since JDK 1.7
	 */
    public List<String> generateParenthesis(int n) {
        List<String> validParen = new ArrayList<String>();
		List<Integer> numberOfLeftParen = new ArrayList<Integer>();
		if(n > 0){
			validParen.add("(");
			numberOfLeftParen.add(1);
			for(int i = 2; i <= 2*n; ++i){
				List<String> tempValidParen = new ArrayList<String>();
				List<Integer> num = new ArrayList<Integer>();
				int length = validParen.size();
				for(int j = 0; j < length; ++j){
					//the length of string in list now is i - 1
					int leftParen = numberOfLeftParen.get(j); // it implies that number of ")" is i - 1 - leftParen					
					String s = validParen.get(j);
					if(leftParen == n){
						tempValidParen.add(s + ")");
						num.add(leftParen);
					}
					else if(leftParen <= (i - 1 - leftParen)){
						tempValidParen.add(s + "(");
						num.add(leftParen + 1);
					}
					else{
						tempValidParen.add(s + "(");
						num.add(leftParen + 1);
						tempValidParen.add(s + ")");
						num.add(leftParen);
					}
				}
				validParen = tempValidParen;
				numberOfLeftParen = num;
			}
		}
		return validParen;
    }
}