LeetCode

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:

// 1AC, simple recursion will do, but keep your mind clear or it's easy to get things complicated
class Solution {
public:
    vector<string> generateParenthesis(int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        result.clear();
        
        dfs(n, n, "");
        
        return result;
    }
private:
    vector<string> result;
    
    void dfs(int cl, int cr, string pat) {
        if(cl == 0 && cr == 0){
            result.push_back(pat);
        }
        
        int i;
        if(cl < cr){
            if(cl > 0){
                dfs(cl - 1, cr, pat + "(");
            }
            if(cr > 0){
                dfs(cl, cr - 1, pat + ")");
            }
        }else if(cl == cr){
            if(cl > 0){
                dfs(cl - 1, cr, pat + "(");
            }
        }else{
            return;
        }
    }
};