Remove Invalid Parentheses(301/Hard)

Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.

Note: The input string may contain letters other than the parentheses ( and ).

例子：
"()())()" -> ["()()()", "(())()"]

"(a)())()" -> ["(a)()()", "(a())()"]

")(" -> [""]

解法1：BFS

public List<String> removeInvalidParentheses(String s) {
      List<String> res = new ArrayList<>();

      // sanity check
      if (s == null) return res;

      Set<String> visited = new HashSet<>();
      Queue<String> queue = new LinkedList<>();

      // initialize
      queue.add(s);
      visited.add(s);

      boolean found = false;

      while (!queue.isEmpty()) {
        s = queue.poll();

        if (isValid(s)) {
          // found an answer, add to the result
          res.add(s);
          found = true;
        }

        if (found) continue;

        // generate all possible states
        for (int i = 0; i < s.length(); i++) {
          // we only try to remove left or right paren
          if (s.charAt(i) != '(' && s.charAt(i) != ')') continue;

          String t = s.substring(0, i) + s.substring(i + 1);

          if (!visited.contains(t)) {
            // for each state, if it's not visited, add it to the queue
            queue.add(t);
            visited.add(t);
          }
        }
      }

      return res;
    }

    // helper function checks if string s contains valid parantheses
    boolean isValid(String s) {
      int count = 0;

      for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (c == '(') count++;
        if (c == ')' && count-- == 0) return false;
      }

      return count == 0;
    }

解法2: DFS+Backtracking

统计左右括号能删的个数，进行DFS。

public List<String> removeInvalidParentheses(String s) {
        Set<String> res = new HashSet<>();
        int rmL = 0, rmR = 0;
        for(int i = 0; i < s.length(); i++) {
            if(s.charAt(i) == '(') rmL++;
            if(s.charAt(i) == ')') {
                if(rmL != 0) rmL--;
                else rmR++;
            }
        }

        DFS(res, s, 0, rmL, rmR, 0, new StringBuilder());
        return new ArrayList<String>(res);
    }

    public void DFS(Set<String> res, String s, int i, int rmL, int rmR, int open, StringBuilder sb) {
        if(i == s.length() && rmL == 0 && rmR == 0 && open == 0) {
            res.add(sb.toString());
            return;
        }
        if(i == s.length() || rmL < 0 || rmR < 0 || open < 0) return;

        char c = s.charAt(i);
        int len = sb.length();

        if(c == '(') {
            DFS(res, s, i + 1, rmL - 1, rmR, open, sb);
            DFS(res, s, i + 1, rmL, rmR, open + 1, sb.append(c));

        } else if(c == ')') {
            DFS(res, s, i + 1, rmL, rmR - 1, open, sb);
            DFS(res, s, i + 1, rmL, rmR, open - 1, sb.append(c));

        } else {
            DFS(res, s, i + 1, rmL, rmR, open, sb.append(c));
        }

        //回溯
        sb.setLength(len);
    }