LeetCode-1249. 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​​.​​

​​题解：​​

​​找出需要删除的括号位置即可。​​

class Solution {
public:
    string minRemoveToMakeValid(string s) {
        int n = s.length();
        string res;
        stack<int> st;
        unordered_set<int> del;
        for (int i = 0; i < n; i++) {
            if (s[i] == '(') {
                st.push(i);
            }
            if (s[i] == ')') {
                if (st.empty() == false) {
                    st.pop();
                }
                else {
                    del.insert(i);
                }
            }
        }
        while (st.empty() == false) {
            del.insert(st.top());
            st.pop();
        }
        for (int i = 0; i < n; i++) {
            if (del.find(i) != del.end()) {
                continue;
            }
            res += s[i];
        }
        return res;
    }
};