LeetCode -- Remove Invalid Parentheses

LeetCode -- Remove Invalid Parentheses
Description:
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 in letters other than the parentheses (and ).


Examples:
()-> [(), ()]
(A) ()-> [(a) (), (a ()]
) (-> []




Input a string containing '(' and ')' and letters. Delete this string for the minimum number of times. If this string is deleted only once, the result is: str1 (The I bit is deleted), str2 (the j bit is deleted)..., strN.
Add the deleted strings to the result set that match the brackets (except for letters, '(' and ')' constitute a valid string.




Ideas:
1. You can use DFS or BFS for this question, but BFS is suitable because it is the minimum number of times rather than the first matching result.
2. if s does not meet the requirements, remove the I-th digit of s, and obtain the new strings t1, t2, t3... judge t [1... n] whether it meets the requirements. if it meets the requirements, add it to the result set in sequence; otherwise, t [0... n) join the queue.
3. The dictionary can be used for pruning because t has been judged during traversal.


Implementation Code:

public class Solution {    public IList
 
   RemoveInvalidParentheses(string s)     {        var ret = new List
  
   ();    var visited = new Dictionary
   
    ();        var q = new Queue
    
     ();    q.Enqueue(s);        var next = new Queue
     
      ();    while(q.Count > 0){    var str = q.Dequeue();    if(IsValid(str))    {    ret.Add(str);    }    else    {    for(var i = 0;i < str.Length; i++){    if(str[i] != '(' && str[i] != ')') continue;        var t = str.Substring(0 , i) + str.Substring(i+1, str.Length - i - 1);    if(!visited.ContainsKey(t))    {    visited.Add(t , 1);    next.Enqueue(t);    }    }    }        if(q.Count == 0){    if(ret.Count > 0){    break;    }        q = new Queue
      
       (next); next.Clear(); } } if(ret.Count == 0){ ret.Add(); } return ret; }private bool IsValid(string s){var stack = new Stack
       
        ();for(var i = 0;i < s.Length; i++){if(s[i] == '('){stack.Push(s[i]);}else if(s[i] == ')'){if(stack.Count == 0){return false;}stack.Pop();}}return stack.Count == 0;}}