Minimum Add to Make Parentheses Valid

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: "())"
Output: 1
Example 2:

Input: "((("
Output: 3
Example 3:

Input: "()"
Output: 0
Example 4:

Input: "()))(("
Output: 4
 

Note:

S.length <= 1000
S only consists of '(' and ')' characters.

题意：只有左右括号的字符串，最少添加多少字符才能完全匹配。
解法一：使用栈依次判断。

class Solution {
public:
    int minAddToMakeValid(string S) {
        stack bracket;
        for(auto s : S){
            if(s == '(' || bracket.empty()) bracket.push(s);
            else if(!bracket.empty()){
                if(bracket.top() == '(')
                    bracket.pop();
                else
                    bracket.push(s);
            }
        }
        return bracket.size();
    }
};

解法二：直接依次判断，思想类似于摩尔投票法。

class Solution {
public:
    int minAddToMakeValid(string S) {
        int left = 0, right = 0;
        for(auto s : S){
            if(s == ')'){
                if(left == 0)
                    right++;
                else
                    left--;
            }
            else
                left++;
        }
        return left + right;
    }
};