add prob #1190; O(N^2) in time and O(N) in space;

Author: randomwalk10

Diff for: 1190_Reverse_Substrings_Between_Each_Pair_of_Parentheses.cpp

@@ -0,0 +1,87 @@
+/*
+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 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.
+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.
+
+来源：力扣（LeetCode）
+链接：https://leetcode-cn.com/problems/reverse-substrings-between-each-pair-of-parentheses
+著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
+*/
+
+/* O(N^2) in time and O(N) in space */
+#include <string>
+#include <stack>
+#include <queue>
+using namespace std;
+
+class Solution {
+public:
+    string reverseParentheses(string s) {
+        stack<char> a;
+        string res;
+
+        for(int i=0; i<(int)s.size(); ++i){
+            if(s[i]=='('){
+                a.push(s[i]);
+            }
+            else if(s[i]==')'){
+                queue<char> b;
+                while(a.top()!='('){
+                    b.push(a.top());
+                    a.pop();
+                }
+                a.pop(); // remove '('
+
+                while(!b.empty()){
+                    a.push(b.front());
+                    b.pop();
+                }
+            }
+            else{
+                a.push(s[i]);
+            }
+        }
+
+        stack<char> b;
+        while(!a.empty()){
+            b.push(a.top());
+            a.pop();
+        }
+        while(!b.empty()){
+            res.push_back(b.top());
+            b.pop();
+        }
+
+        return res;
+    }
+};