1190 Reverse Substrings Between Each Pair of Parentheses 反转每对括号间的子串

Description:
You are 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 brackets.

Example:

Example 1:

Input: s = "(abcd)"
Output: "dcba"

Example 2:

Input: s = "(u(love)i)"
Output: "iloveu"
Explanation: The substring "love" is reversed first, then the whole string is reversed.

Example 3:

Input: s = "(ed(et(oc))el)"
Output: "leetcode"
Explanation: First, we reverse the substring "oc", then "etco", and finally, the whole string.

Constraints:

1 <= s.length <= 2000
s only contains lower case English characters and parentheses.
It is guaranteed that all parentheses are balanced.

题目描述:
给出一个字符串 s（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。

示例 :

示例 1：

输入：s = "(abcd)"
输出："dcba"

示例 2：

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。

示例 3：

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。

示例 4：

输入：s = "a(bcdefghijkl(mno)p)q"
输出："apmnolkjihgfedcbq"

提示:

0 <= s.length <= 2000
s 中只有小写英文字母和括号
题目测试用例确保所有括号都是成对出现的

思路:

1. 栈简单模拟
   用栈记录每次左括号出现的位置
   每次遇到匹配的右括号就将括号内的所有字符反转
   时间复杂度为 O(n ^ 2), 空间复杂度为 O(n)
2. 栈线性模拟
   同时记录两个括号出现的位置, 然后每一对括号实际上只用记录对应方向
   每次碰到括号就反向将字符加入结果中
   时间复杂度为 O(n), 空间复杂度为 O(n)

代码:
C++:

class Solution 
{
public:
    string reverseParentheses(string s) 
    {
        stack<int> s_stack;
        for (int i = 0, n = s.size(); i < n; i++) 
        {
            if (s[i] == '(') s_stack.push(i);
            if (s[i] == ')') 
            {
                reverse(s.begin() + s_stack.top() + 1, s.begin() + i);
                s_stack.pop();
            }
        }
        s.erase(remove(s.begin(), s.end(), '('), s.end());
        s.erase(remove(s.begin(), s.end(), ')'), s.end());
        return s;
    }
};

Java:

class Solution {
    public String reverseParentheses(String s) {
        StringBuilder sb = new StringBuilder();
        Stack<Integer> stack = new Stack<>();
        int n = s.length(), pair[] = new int[n], index = 0, step = 1;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '(') stack.push(i);
            else if (s.charAt(i) == ')') {
                int j = stack.pop();
                pair[i] = j;
                pair[j] = i;
            }
        }
        while (index < n) {
            if (s.charAt(index) == '(' || s.charAt(index) == ')') {
                index = pair[index];
                step = -step;
            } else sb.append(s.charAt(index));
            index += step;
        }
        return sb.toString();
    }
}

Python:

class Solution:
    def reverseParentheses(self, s):
        return s if '(' not in s else self.reverseParentheses(re.sub(r'\(([^()]*)\)', lambda x: x.group(1)[::-1], s))