1298. Reverse Substrings Between Each Pair of Parentheses

• Difficulty: Medium

• Topics: Stack

• Similar Questions:

Problem:

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.

Solutions:

class Solution {
public:
    string reverseParentheses(string s) {
        int pos = 0;
        return helper(s, pos);
    }

private:
    string helper(const string& s, int& pos) {
        string ret;
        while (pos < s.length()) {
            if (s[pos] == '(') {
                string inner = helper(s, ++pos);
                reverse(inner.begin(), inner.end());
                ret.append(inner);
            } else if (s[pos] == ')') {
                ++pos;
                return ret;
            } else {
                ret.push_back(s[pos++]);
            }
        }
        return ret;
    }

};