Leetcode_1249_Minimum Remove to Make Valid Parentheses

[lihe]

Minimum Remove to Make Valid Parentheses

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.

算法思路：利用栈来解决这题，栈中存储下标。从左到有遍历字符串，如果先出现)，则此位置的)直接不可用；如果出现(，则将下标压进栈中，继续遍历；再遇到)，则查看栈中是否有(，如果有则将其弹出，否则，此位置的)不可用。

class Solution {
    public String minRemoveToMakeValid(String s) {
        Stack<Integer> stc = new Stack<>();
        boolean[] invalidIndex = new boolean[s.length()];

        StringBuilder result = new StringBuilder();
        for(int i = 0; i < s.length(); i++){
            if(s.charAt(i) == '('){
                stc.push(i);
                invalidIndex[i] = true;
            }

            if(s.charAt(i) == ')'){
                if(stc.empty()){
                    invalidIndex[i] = true;
                }else {
                    invalidIndex[stc.pop()] = false;
                }
            }
        }
        for(int i = 0; i < s.length(); i++){
            if(!invalidIndex[i]){
                result.append(s.charAt(i));
            }
        }
        return result.toString();
    }
}