[LeetCode]22 Generate Parentheses

public class Solution {

public List<String> generateParenthesis(int n) {

// Solution B:
// return generateParenthesis_BruteForce(n);

// Solution A:
return generateParenthesis_Dynamic(n);
}

////////////////////////////////////////
// Solution A: Dynamic
//
public List<String> generateParenthesis_Dynamic(int n)
{
List<String> result = new ArrayList<>();
generate(n, 0, 0, "", result);
return result;
}

private void generate(int n, int left, int right, String s, List<String> result)
{
if (right > left)
return; // ) cannot be more than (

if (left == n && right == n)
{
// A valid result.
result.add(s);
return;
}

if (left == n) // right ')' is less than n
{
generate(n, left, right + 1, s + ")", result);
return;
}

generate(n, left + 1, right, s + "(", result); // Add (
generate(n, left, right + 1, s + ")", result); // Add )
}

////////////////////////////////////////
// Solution A: Brute Force
//
public List<String> generateParenthesis_BruteForce(int n) {

if (n < 0)
return null;    // Invalid input

if (n == 0)
return Collections.<String> emptyList();

List<Set<String>> results = new ArrayList<>(n);
results.add(Collections.<String> singleton("()"));  // Result for 1;

for (int i = 1 ; i < n ; i ++)
{
Set<String> curR = new HashSet<>();

Set<String> lastR = results.get(i - 1);
for (String s : lastR)
{
// Add one more '(' in the begining;
s = "(" + s;
// Iterate s, Add ')' after each '('
for (int j = 0 ; j < s.length() ; j ++)
{
if (s.charAt(j) == '(')
{
String newStr = s.substring(0, j + 1) + ")" + s.substring(j + 1, s.length());
curR.add(newStr);
}
}
}

results.add(curR);
}
return new ArrayList<>(results.get(n - 1));
}
}