[LeetCode] 921. Minimum Add to Make Parentheses Valid

Description

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:

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

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Analyse

找出能使括号字符串合法的最小括号数

用栈当然可以搞定，看最后栈里剩下的元素个数就是要求的数

直接遍历一遍字符串，数(，)的个数，最开始我的做法是直接)的个数-(的个数，仔细一想不对，()))((这个用例就通不过，括号的顺序很重要

从左到右遍历字符串

1. 遇到(就计数器+1
2. 遇到)，如果之前还有(就直接消去一个(，否则就要增加一个(来抵消这个)

Code

int minAddToMakeValid(string S)
{
    int left_count = 0;
    int total = 0;

    for(char c : S)
    {
        if (c == '(')
        {
            ++left_count;
            ++total;
        }
        else
        {
            if (left_count > 0)
            {
                --left_count;
                --total;
            }
            else
            {
                ++total;
            }
        }
    }

    return total;
}

Result

4ms faster than 100.00%

Others

吐槽下改版后LeetCode的界面真难看，三层额头，看题的空间都变小了