Generate Parentheses

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:

"((()))", "(()())", "(())()", "()(())", "()()()"

解答：

我的方法：回溯

class Solution {
private:
vector<string> vec;
int num = 0;
public:
    void parenthesis(string str, int i, int j, int number){
    if(number == num && i == num && j == num){
        vec.push_back(str);
        return;
    }
    if(i >= num+1 || j>=num+1)
        return;
        parenthesis(str+"(",i+1,j,number);
    if(j<i)
        parenthesis(str+")",i,j+1,number+1);
}

vector<string> generateParenthesis(int n) {
    string s;
    num = n;
    int i = 0, j = 0,number = 0;
    parenthesis(s,i,j,number);
    return vec;
}
};

2.跄跄~~大牛的算法 用动态规划

My method is DP. First consider how to get the result f(n) from previous result f(0)...f(n-1). Actually, the result f(n) will be put an extra () pair to f(n-1). Let the "(" always at the first position, to produce a valid result, we can only put ")" in a way that there will be i pairs () inside the extra () and n - 1 - i pairs () outside the extra pair.

Let us consider an example to get clear view:

f(0): ""

f(1): "("f(0)")"

f(2): "("f(0)")"f(1), "("f(1)")"

f(3): "("f(0)")"f(2), "("f(1)")"f(1), "("f(2)")"

So f(n) = "("f(0)")"f(n-1) , "("f(1)")"f(n-2) "("f(2)")"f(n-3) ... "("f(i)")"f(n-1-i) ... "(f(n-1)")"

Below is my code:

public class Solution
{
    public List<String> generateParenthesis(int n)
    {
        List<List<String>> lists = new ArrayList<>();
        lists.add(Collections.singletonList(""));

        for (int i = 1; i <= n; ++i)
        {
            final List<String> list = new ArrayList<>();

            for (int j = 0; j < i; ++j)
            {
                for (final String first : lists.get(j))
                {
                    for (final String second : lists.get(i - 1 - j))
                    {
                        list.add("(" + first + ")" + second);
                    }
                }
            }

            lists.add(list);
        }

        return lists.get(lists.size() - 1);
    }
}