921. Minimum Add to Make Parentheses Valid

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

Example 3:

Input: s = "()"
Output: 0

Example 4:

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

Constraints:

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

思路：与20. Valid Parentheses类似，是关于字符串匹配的问题。

想法之一就是利用堆栈。设置整数型sum，表示最少添加的括号数目。如果是左括号，压入堆栈；如果是右括号，查看堆栈栈顶元素，如果是左括号，那就证明该右括号有对应的匹配项，如果没有，（没有左括号，更不会有右括号！！），那就sum++；

堆栈中就只有左括号，又因为只有一个类型的括号匹配！一定需要用堆栈吗？不一定

该栈的作用，其实，相当于一个计数器。故上述思路就翻译成：设置一个整数型sum，表示最少添加的括号数目；设置整数型counter，表示假想stack里面的左括号数目。如果是左括号，counter++; 如果是右括号，查看counter的值，若>0，则该右括号一定有匹配的对应的左括号，则有counter--；如果=0，则证明假想stack里面，没有左括号啦。所以就要sum++

代码如下：

    public static int minAddToMakeValid(String s) {
        int sum = 0;
        int count = 0;
        if (s == null) return 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i)=='(') {
                count++;
            } else {
                if (count<=0) {
                    sum = sum + 1;
                } else {
                    count--;
                }
            }
        }
        return sum + count;
    }