Given a string s of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid.

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

• It is the empty string, 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.

Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.

 

Example 1:

Input: s = "())"
Output: 1

Example 2:

Input: s = "((("
Output: 3

Example 3:

Input: s = "()"
Output: 0

Example 4:

Input: s = "()))(("
Output: 4

 

Note:

1. s.length <= 1000
2. s only consists of '(' and ')' characters.

题目链接：https://leetcode.com/problems/minimum-add-to-make-parentheses-valid/

题目大意：最少添加几个括号可以使之合法

题目分析：栈模拟即可

class Solution {
    public int minAddToMakeValid(String s) {
        if (s.length() < 2) {
            return s.length();
        }
        Stack<Character> stk = new Stack<>();
        char[] str = s.toCharArray();
        int i = 0, ans = 0;
        while (i < str.length) {
            if (str[i] == ')') {
                if (!stk.empty()) {
                    stk.pop();
                } else {
                    ans++;
                }
            } else {
                stk.add(str[i]);
            }
            i++;
        }
        ans += stk.size();
        return ans;
    }
}

不难发现这个栈其实也没啥用，只是用来记个数，所以可以省略

0ms，时间击败100%

class Solution {
    public int minAddToMakeValid(String s) {
        if (s.length() < 2) {
            return s.length();
        }
        char[] str = s.toCharArray();
        int ans = 0, stk = 0;
        for (int i = 0; i < str.length; i++) {
            if (str[i] == ')') {
                if (stk != 0) {
                    stk--;
                } else {
                    ans++;
                }
            } else {
                stk++;
            }
        }
        ans += stk;
        return ans;
    }
}