[Leetcode 1021] Remove Outermost Parentheses

原题说明

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:

1. S.length <= 10000
2. S[i] is "(" or ")"
3. S is a valid parentheses string

解题思路

用一个整型变量opened来记录左括号比右括号多出的数量
遍历S，遇到左括号++opened，遇到右括号--opened
除非遇到第一个左括号（opened = 1）或者最后一个右括号（opened = 0）
不然其余字符均加入最终结果res

示例代码 (cpp)

class Solution {
public:
    string removeOuterParentheses(string S) {
        int opened = 0;
        string res;
        for (auto ch : S) {
            if (ch == '(') {
                ++opened;
                if (opened != 1) {
                    res += ch;
                }
            } else {
                --opened;
                if (opened != 0) {
                    res += ch;
                }
            }
        }
        return res;
    }
};

示例代码（java）

class Solution {  
    public String removeOuterParentheses(String S) {
        StringBuilder s = new StringBuilder();
        int opened = 0;
        for (char c : S.toCharArray()) {
            if (c == '(' && opened++ > 0) s.append(c);
            if (c == ')' && opened-- > 1) s.append(c);
        }
        return s.toString();
    }
}

示例代码（python）

class Solution(object):
    def removeOuterParentheses(self, S):
        """
        :type S: str
        :rtype: str
        """
        res, opened = [], 0
        for c in S:
            if c == '(' and opened > 0: res.append(c)
            if c == ')' and opened > 1: res.append(c)
            opened += 1 if c == '(' else -1
        return "".join(res)

复杂度分析

时间复杂度: O(N)
空间复杂度: O(N) 如果包含返回的string，不然为O(1)

归纳总结

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