【leetcode】1190. Reverse Substrings Between Each Pair of Parentheses

题目如下：

Given a string s that consists of lower case English letters and brackets.

Reverse the strings in each pair of matching parentheses, starting from the innermost one.

Your result should not contain any bracket.

Example 1:
Input: s = "(abcd)"

Output: "dcba"
Example 2:
Input: s = "(u(love)i)"

Output: "iloveu"
Example 3:
Input: s = "(ed(et(oc))el)"

Output: "leetcode"
Example 4:
Input: s = "a(bcdefghijkl(mno)p)q"

Output: "apmnolkjihgfedcbq"
Constraints:

0 <= s.length <= 2000

s only contains lower case English characters and parentheses.

It's guaranteed that all parentheses are balanced.

解题思路：本题和leetcode之前出现过的四则运算的题目类似。从头开始遍历s，不是'('的字符直接入栈，如果遇到')'，找出栈中最靠近栈顶的'('，逆置从'('到栈顶的所有元素，同时删除'('，直到s遍历完成为止。

代码如下：

class Solution(object):

    def reverseParentheses(self, s):

        """

        :type s: str

        :rtype: str

        """

        left_inx = []

        stack = []

        for i in s:

            if i == '(':

                stack.append(i)

                left_inx.append(len(stack)-1)

            elif i == ')':

                inx = left_inx.pop(-1)

                sub = stack[inx + 1:]

                sub.reverse()

                stack = stack[:inx] + sub

            else:

                stack.append(i)

        return ''.join(stack)

巴特西

【leetcode】1190. Reverse Substrings Between Each Pair of Parentheses

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