022 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:

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

我的解法很复杂，不知道怎么搞的，其实思路很简单，使用的是全排序思想，一个位置上只有两种状态，然后使用经典的方法就行

        public List<String> generateParenthesis(int n) {

		LinkedList<Character> result=new LinkedList<>();
		Set<String> set=new HashSet<>();
		List<String> list=new ArrayList<>();
		generateLoop(n,n,result,set);
		for (String string : set) {
			list.add(string);
		}
		return list;
	}
	
	public void generateLoop(int r,int l,LinkedList<Character> result,Set<String> list){
		if(r==0||l==0){
			if(l==0){
				checkStack(result,r,list);
			}
			return;
		}
		result.push('(');
		l=l-1;
		generateLoop(r, l, result, list);
		
		result.pop();
		l=l+1;
		result.push(')');
		r=r-1;
		generateLoop(r, l, result, list);
		result.pop();
		r=r+1;
	}

	private void checkStack(LinkedList<Character> result, int r,Set<String> list) {
		String str="";
		for(int i=0;i<result.size();i++)str+=result.get(i);
		for(int i=0;i<r;i++)str+=")";
		LinkedList<Character> reserve=new LinkedList<>();
		for(int i=0;i<str.length();i++){
			if(str.charAt(i)=='('){
				reserve.push(str.charAt(i));
			}else{
				if(reserve.isEmpty()){
					return;
				}
				if(reserve.pop()!='(')return;
			}
		}
		list.add(str);
		
	}