LeetCode-Remove Outermost Parentheses

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Description：
A valid parentheses string is either empty (""), “(” + A + “)”, or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, “”, “()”, “(())()”, and “(()(()))” are all valid parentheses strings.

A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string S, consider its primitive decomposition: $S = P_1 + P_2 + ... + P_k$, where $P_i$ are primitive valid parentheses strings.

Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.

Example 1:

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

Example 2:

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

Example 3:

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

Note:

• S.length <= 10000
• S[i] is “(” or “)”
• S is a valid parentheses string

题意：给定一个字符串S为一个合法的括号组合，将S划分为 $S = P_1 + P_2 + ... + P_k$，其中 $P_1,P_2,..., P_k$也为合法的括号组合并且是不可再分的；去除 $P_1,P_2,..., P_k$最外层括号并连接后，作为最后的结果返回；

解法：在括号匹配的问题中我们可以利用栈来判断是否合法，同样的在这个问题中，我们可以利用栈来找到每一个最小的不可再分的 $P_1,P_2,..., P_k$；流程如下

1. 将第一个字符添加到stack中
2. 遍历字符串直到栈为空
3. 遇到左括号则添加到栈中，右括号则删除栈顶的元素
4. 栈空时，遍历过的字符串位置即为一个最小不可再分的 $P_1,P_2,..., P_k$

Java

class Solution {
    public String removeOuterParentheses(String S) {
        StringBuilder res = new StringBuilder();
        LinkedList<Character> stack = new LinkedList<>();
        for (int st = 0, ed = 0; ed != S.length();) {
            st = ed;
            stack.addLast(S.charAt(ed));
            ed++;
            while (stack.size() != 0 && ed != S.length()) {
                char ch = S.charAt(ed);
                if (ch == '(') {
                    stack.addLast(ch);
                } else {
                    stack.pollLast();
                }
                ed++;
            }
            res.append(S.substring(st + 1, ed - 1));
        }
        
        return res.toString();
    }
}