LeetCode - Generate Parentheses

Generate Parentheses
2013.12.7 00:49
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: "((()))", "(()())", "(())()", "()(())", "()()()"
Solution:
　　If you're familiar with the Catalan Numbers , you'll see it's a typical solution for this problem.
　　Since all the combinations are required, we'll use DFS to go through every possible sequence.
　　Let's consider the recursion in the following way:
　　　　1. For a sequence of length 2 * n, there're n '(' and n ')'
　　　　2. For a valid sequence, number of '(' will never be less than the number of ')' at any position. If there're more ')' than '(', there must be a mismatch.
　　　　3. Always do the recursion from left to right.
　　Time complexity is H(n) * O(n) = C(2 * n, n)/(n + 1) * n, that's roughly O(n!). Space complexity is not sure yet...
Accepted code:

 1 // 1AC, simple recursion will do, but keep your mind clear or it's easy to get things complicated

 2 class Solution {

 3 public:

 4     vector<string> generateParenthesis(int n) {

 5         // IMPORTANT: Please reset any member data you declared, as

 6         // the same Solution instance will be reused for each test case.

 7         result.clear();

 8         

 9         dfs(n, n, "");

10         

11         return result;

12     }

13 private:

14     vector<string> result;

15     

16     void dfs(int cl, int cr, string pat) {

17         if(cl == 0 && cr == 0){

18             result.push_back(pat);

19         }

20         

21         int i;

22         if(cl < cr){

23             if(cl > 0){

24                 dfs(cl - 1, cr, pat + "(");

25             }

26             if(cr > 0){

27                 dfs(cl, cr - 1, pat + ")");

28             }

29         }else if(cl == cr){

30             if(cl > 0){

31                 dfs(cl - 1, cr, pat + "(");

32             }

33         }else{

34             return;

35         }

36     }

37 };