題目說明:給你一組括號字串，要你移除掉最外部的括號，由於詳細的題目敘述實在太長，這邊直接省略。

Case 1
Input: s = "(()())(())"
Output: "()()()"
Explanation:
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Case 2:
Input: s = "(()())(())(()(()))"
Output: "()()()()(())"
Explanation:
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".

Case 3:
Input: s = "()()"
Output: ""
Explanation:
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".

解題思路:這題同樣是使用stack，將左括號push，遇到右括號要pop。但是在結果的呈現上因為最外部的括號省略，因此我們要先判斷當前stack的size是否大於0，大於0才要需要將左括號加進去結果裡面，如果等於0就代表是最外部的括號;說了這麼多可能有些聽起來還是有聽沒有懂，因此直接用例子來進行說明或許變得簡單明瞭。

(()())這組字串當中，一開始遇到左括號，stack要push進去，但是在結果顯示上我們不需要這個括號因為它是最外部的括號。因此在push到stack之前我們要先考慮到這個stack的size是否>0，進而進行判斷是否為最外部的括號。同時對於右括號結果上的處理也是同樣的邏輯，步驟如下:

Step Stack Result
0 ( ""
1 (( "("
2 ( "()"
3 (( "()("
4 ( "()()"
5 empty "()()"

附上程式碼和註解

Java

class Solution {
    public String removeOuterParentheses(String s) {
        Stack<Character>st=new Stack<>();
        StringBuilder sb=new StringBuilder();
        for(int i=0;i<s.length();i++)
        {
            char ch=s.charAt(i);
            if(ch=='(')
            {
                if(st.size()>0)
                {
                    sb.append(ch);
                }
                st.push(ch);
            }else
            {
                st.pop();
                if(st.size()>0)
                {
                    sb.append(ch);
                }
            }
        }
        return sb.toString();
    }
}

Python

class Solution:
    def removeOuterParentheses(self, s: str) -> str:
        res=''
        stack=[]
        for i in s:
            if i=='(':
                if len(stack)>0:
                    res+=i
                stack.append(i)
            else:
                stack.pop()
                if len(stack)>0:
                    res+=i
        return res