LeetCode301. Remove Invalid Parentheses

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 ).

Example 1:

Input: "()())()"
Output: ["()()()", "(())()"]

Example 2:

Input: "(a)())()"
Output: ["(a)()()", "(a())()"]

Example 3:

Input: ")("
Output: [""]

思路：可以利用DFS或者BFS来解这道题，感觉还是BFS简单点，即对于从给定的字符串通过移除 ( 或 ) 来构造所有可能的合法串，如果合法就加入到set集合中，不合法到到下一轮的BFS中。

public class Solution {
    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();
        
　　　　// 如果当前层次中有合法解的话，只需要将当前层次中的字符串全部弹出判断是否合法，停止BFS，这样保证所得到的合法字符串是移除最少字符得到的
        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;
    }
}