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:

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

完全想不到，在看了別人的演算法後，自己畫圖加以理解才明白，主要是利用兩個條件去限制迭代，把不合格的排除掉，第一個open < max的條件篩選掉的用感嘆號表示，第二個條件close < open

程式碼如下：

class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<>();
        StringBuilder sb = new StringBuilder();
        helper(res, sb, 0
 
, 0, n);
        return res;
    }
    public void helper(List<String> res, StringBuilder sb, int open, int close, int max){
        if(open == max && close == max){
            res.add(sb.toString());
            return;
        }

        if(open < max){
            sb.append("(");
            helper(res, sb, open
 
+1, close, max);
            sb.setLength(sb.length()-1);
        }
        if(close < open){
            sb.append(")");
            helper(res, sb, open, close+1, max);
            sb.setLength(sb.length()-1);
        }
    }
}