题目：

A parentheses string is valid if and only if:

• It is the empty string,
• 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.

You are given a parentheses string s. In one move, you can insert a parenthesis at any position of the string.

• For example, if s = "()))", you can insert an opening parenthesis to be "(()))" or a closing parenthesis to be "())))".

Return the minimum number of moves required to make s valid.

Example 1:

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

Example 2:

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

Constraints:

• 1 <= s.length <= 1000
• s[i] is either '(' or ')'.

解法：

class Solution {
    public int minAddToMakeValid(String s) {
        int result = 0, bal = 0;
        for(int i = 0; i < s.length(); i++){
            bal += s.charAt(i) == '(' ? 1 : -1;
            if(bal == -1){
                bal = 0;
                result++;
            }
        }
        return bal+result;
    }
}

从左遍历，记录多余的右括号和未匹配的左括号