Leetcode/各大家 -- 22. Generate Parentheses(Backtracking)

22. Generate Parentheses(Backtracking)
medium

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:

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

 Google Uber Zenefits

思路： - use dfs, every point, it has two options '(', ')'. We have n left bracket '(', n right bracket ')',    
            - when both  left and right are used up, we add to result.
            - Watch return cases! if left > right, no valid answer. eg: )(())(
               Also we can't have > n left or right bracket
Complexity: Time O(2^n)  Space O(n)

public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<String>();
        dfs(new StringBuilder(), n, n,res);
        return res;
    }
    public void dfs(StringBuilder path,int left,int right,List<String> res){
        int len = path.length();
        if(left > right || left < 0 || right < 0){
            return;
        }
        if(left == 0 && right == 0){
            res.add(path.toString());
            return;
        }
        dfs(path.append("("),left-1,right,res);
        path.setLength(len);
        dfs(path.append(")"),left,right-1,res);
        path.setLength(len);
    }