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.

一看，还以为是301 Remove Invalid Parentheses呢。其实这我就用20 Valid Parentheses那题的办法。把要remove的括号放到deque里。一开始是用栈的，因为我们要返回remove了之后的字符串。所以用deque方便点。T:O(n), S:O(n)

public String minRemoveToMakeValid(String s) {
    if (s == null || s.isEmpty()) {
        return "";
    }

    // parenthesis's location
    Deque<Integer> deque = new LinkedList<>();
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (c == ')' && !deque.isEmpty() && s.charAt(deque.peekLast()) == '(') {
            deque.removeLast();
        } else if (c == '(' || c == ')') {
            deque.addLast(i);
        }
    }

    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < s.length(); i++) {            
        if (!deque.isEmpty() && i == deque.getFirst()) {
            deque.removeFirst();
            continue;
        }
        sb.append(s.charAt(i));
    }

    return sb.toString();
}