1021. Remove Outermost Parentheses

LeetCode 原题

A valid parentheses string is either empty "", "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation.

• For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.

A valid parentheses string s is primitive if it is nonempty, and there does not exist a way to split it into s = A + B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string s, consider its primitive decomposition: s = P1 + P2 + ... + Pk, where Pi are primitive valid parentheses strings.

Return s after removing the outermost parentheses of every primitive string in the primitive decomposition of s.

Constraints:

• 1 <= s.length <= 105
• s[i] is either '(' or ')'.
• s is a valid parentheses string.

思路：看图写话

这题的意思很明确：删除所有深度为0的括号。那么只要遍历一遍 排除掉他们就可以了

标答给了两种方法 实际上是等价的。栈方法是用栈是否为空来标记当前层数是否为0

为何右括号要写在深度判断的前面，深度判断要写在左括号的前面？因为左括号真正的深度产生于进入左括号之前；右括号的真正的深度产生于离开右括号之后。

class Solution:
    def removeOuterParentheses(self, s: str) -> str:
        dph = 0
        ans = ''
        for c in s:
            if c == ')':
                dph -= 1
            if dph > 0:
                ans += c
            if c == '(':
                dph += 1
        return ans