leetcode: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).

Examples:

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

Credits:

Special thanks to @hpplayer for adding this problem and creating all test cases.

class Solution(object):        def isValid(self, s):            leftCount = 0        for c in s:            if c == '(':                leftCount = leftCount+1            elif c == ')':                if leftCount > 0:                    leftCount = leftCount-1                else:                    return False                if leftCount == 0:            return True        else:            return False                    def removeInvalidParentheses(self, s):        """        :type s: str        :rtype: List[str]        """                q = [s]        ans = []        visited = set(q)                found = False                while len(q) > 0:            cur = q.pop(0)            if self.isValid(cur):                ans.append(cur)                found = True            elif not found:                for i in xrange(len(cur)):                    if cur[i] == '(' or cur[i] == ')':                        candidate = cur[:i] + cur[i+1:]                                            if candidate not in visited:                            q.append(candidate)                            visited.add(candidate)                            return ans                    

BFS的思想，邻接节点是 比当前节点改变一个字节的节点

https://leetcode.com/discuss/67842/share-my-java-bfs-solution

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();        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;    }}

还有一个DFS思想的解法

https://leetcode.com/discuss/72208/easiest-9ms-java-solution

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));     }    sb.setLength(len);}