22. Generate Parentheses

• Difficulty: Medium

• Topics: String, Backtracking

• Similar Questions:

  • Letter Combinations of a Phone Number

  • Valid Parentheses

Problem:

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:

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

Solutions:

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> ret;
        if (n == 0) return ret;
        string path;

        helper(2 * n, 0, path, ret);
        return ret;
    }

    void helper(int remain, int left, string& path, vector<string>& ret) {
        if (remain == 0 && left == 0) {
            ret.push_back(path);
            return;
        }
        if (left < 0) {
            return;
        }
        if (remain < left) {
            return;
        }

        path.push_back('(');
        helper(remain - 1, left + 1, path, ret);
        path.pop_back();
        path.push_back(')');
        helper(remain - 1, left - 1, path, ret);
        path.pop_back();

    }
};

More intuitive solution

From Huahua

// Author: Huahua
// Running time: 0 ms
class Solution {
public:
  vector<string> generateParenthesis(int n) {
    vector<string> ans;
    string cur;
    if (n > 0) dfs(n, n, cur, ans);
    return ans;
  }
private:
  void dfs(int l, int r, string& s, vector<string>& ans) {
    if (l + r == 0) {
      ans.push_back(s);
      return;
    }
    if (r < l) return;
    if (l > 0) {
      dfs(l - 1, r, s += "(", ans);
      s.pop_back();
    }
    if (r > 0) {
      dfs(l, r - 1, s += ")", ans);
      s.pop_back();
    }
  }
};