LeetCode1249. 移除無效的括號

1249. Minimum Remove to Make Valid Parentheses

[Medium] 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.

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題目：給你一個由 '('、')' 和小寫字母組成的字符串 s。你需要從字符串中刪除最少數目的 '(' 或者 ')' （可以刪除任意位置的括號)，使得剩下的「括號字符串」有效。請返回任意一個合法字符串。

思路：棧。當)數大於(數時，括號對失衡。用bt標記需要刪除的括號。同921. Minimum Add to Make Parentheses Valid。

工程代碼下載 GitHub

class Solution {
public:
    string minRemoveToMakeValid(string s) {
        const int n = s.size();
        vector<bool> bt(n);
        stack<int> sk;
        for(int i = 0; i < s.size(); ++i){
            if(s[i] == '(')
                sk.push(i);
            else if(s[i] == ')'){
                if(sk.empty())
                    bt[i] = true;
                else
                    sk.pop();
            }
        }

        while(!sk.empty()){
            bt[sk.top()] = true;
            sk.pop();
        }

        string res;
        for(int i = 0; i < n; ++i){
            if(bt[i] == false)
                res.push_back(s[i]);
        }

        return res;
    }
};

或者使用佔位符*，以下轉自votrubac。

string minRemoveToMakeValid(string s) {
    stack<int> st;
    for (auto i = 0; i < s.size(); ++i) {
        if (s[i] == '(') st.push(i);
        if (s[i] == ')') {
            if (!st.empty()) st.pop();
            else s[i] = '*';
        }
    }
    while (!st.empty()) {
        s[st.top()] = '*';
        st.pop();
    }
    s.erase(remove(s.begin(), s.end(), '*'), s.end());
    return s;
}