reverse-substrings-between-each-pair-of-parentheses

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Algorithms

You are given a string <code>s</code> 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 brackets.

&nbsp;
Example 1:

<pre>
Input: s = &quot;(abcd)&quot;
Output: &quot;dcba&quot;
</pre>

Example 2:

<pre>
Input: s = &quot;(u(love)i)&quot;
Output: &quot;iloveu&quot;
Explanation: The substring &quot;love&quot; is reversed first, then the whole string is reversed.
</pre>

Example 3:

<pre>
Input: s = &quot;(ed(et(oc))el)&quot;
Output: &quot;leetcode&quot;
Explanation: First, we reverse the substring &quot;oc&quot;, then &quot;etco&quot;, and finally, the whole string.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> only contains lower case English characters and parentheses.</li>
	<li>It is guaranteed that all parentheses are balanced.</li>
</ul>

reverseParentheses

Reverse Substrings Between Each Pair of Parentheses

Stack

String

给出一个字符串&nbsp;<code>s</code>（仅含有小写英文字母和括号）。

请你按照从括号内到外的顺序，逐层反转每对匹配括号中的字符串，并返回最终的结果。

注意，您的结果中 不应 包含任何括号。

&nbsp;

示例 1：

<pre>
输入：s = "(abcd)"
输出："dcba"
</pre>

示例 2：

<pre>
输入：s = "(u(love)i)"
输出："iloveu"
解释：先反转子字符串 "love" ，然后反转整个字符串。</pre>

示例 3：

<pre>
输入：s = "(ed(et(oc))el)"
输出："leetcode"
解释：先反转子字符串 "oc" ，接着反转 "etco" ，然后反转整个字符串。</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> 中只有小写英文字母和括号</li>
	<li>题目测试用例确保所有括号都是成对出现的</li>
</ul>