22. Generate Parentheses

Problem statement:

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:

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

Analysis:

This problem also involves DFS, but is different with the DFS template I have summarized, it belongs to the questions that just given a number and find some possible solutions. It does not belong to the problems which relate to an array or string. This is a new DFS philosophy, but it looks simple.

Some key points of this problem:

All possible solutions: DFS without a return value.

n is the given number: it is the terminate condition of DFS meanwhile we also need to add one possible solution.

I struggled for how to orderly arrange "(" and ")", but when I checked the answer, I found that this is not what we need to care, we need to check that in each loop, the number of "(" should always be greater or equal than the number of ")". This is another terminate condition, but no any action for the answer.

Solution:

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<string> sets;
        generate_parenthesis(sets, "", n, n);
        return sets;
    }
private:
    void generate_parenthesis(vector<string>& sets, string seq, int left, int right){
        // minus 1 each time when a "(" or ")" is add
        // so left should always less than right
        if(left  >  right){
            return;
        }
        if(left == 0 && right == 0){
            sets.push_back(seq);
            return;
        }
        // add "("
        if(left > 0){
            generate_parenthesis(sets, seq + "(", left - 1, right);
        }
        // add ")"
        if(right > 0){
            generate_parenthesis(sets, seq + ")", left, right - 1);
        }
        return;
    }
};