（Leetcode）22. Generate 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:
“((()))”, “(()())”, “(())()”, “()(())”, “()()()”

Solution 1:

Use two integers to count the remaining left parenthesis (n) and the right parenthesis (m) to be added. At each function call add a left parenthesis if n >0 and add a right parenthesis if m>0. Append the result and terminate recursive calls when both m and n are zero.

public List<String> generateParenthesis(int n) {
        LinkedList<String> resultList = new LinkedList<String>();
        generatePar(resultList, "", 0, 0, n);

        return resultList;
    }

    private void generatePar(LinkedList<String> resultList,String str, int pre, int post, int max){
        if(str.length()==max*2){
            resultList.add(str);
            return;
        }

        if(pre<max){
            generatePar(resultList, str+"(", pre+1, post, max);
        }
        if(post<pre){
            generatePar(resultList, str+")", pre, post+1, max);
        }
    }

Solution 2:

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 the 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);
    }
}