1249. Minimum Remove to Make Valid Parentheses ------M

Given a string s of '(' , ')' and lowercase English characters. 

Your task is to remove the minimum number of parentheses ( '(' or ')', in any positions ) so that the resulting parentheses string is valid and return any valid string.

Formally, a parentheses string is valid if and only if:

• It is the empty string, contains only lowercase characters, or
• It can be written as AB (A concatenated with B), where A and B are valid strings, or
• It can be written as (A), where A is a valid string.

 

Example 1:

Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.

Example 2:

Input: s = "a)b(c)d"
Output: "ab(c)d"

Example 3:

Input: s = "))(("
Output: ""
Explanation: An empty string is also valid.

Example 4:

Input: s = "(a(b(c)d)"
Output: "a(b(c)d)"

 

Constraints:

• 1 <= s.length <= 10^5
• s[i] is one of  '(' , ')' and lowercase English letters.

 

A:

就是每次找配对儿的。找不到，则标记为不行

class Solution {
public:
    string minRemoveToMakeValid(string s) {
        int n = s.length();
        vector<bool> deleted(n,false);
        stack<int> S;// stack of (, waiting to be matched.
        for(int i =0;i<n;i++){
            char ch = s[i];
            if(ch == '('){
                S.push(i);
            }else if(ch ==')'){
                if(S.empty()){
                    deleted[i] = true;
                }else{
                    S.pop();
                }
            }
        }
        while(not S.empty()){
            int tmp = S.top();
            S.pop();
            deleted[tmp] = true;
        }
        string res = "";
        for(int i =0;i<n;i++){
            char ch = s[i];
            if(not deleted[i]){
                res += ch;
            }
        }
        return res;
    }
};