1021 Remove Outermost Parentheses 删除最外层的括号
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:
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
题目描述:
有效括号字符串为空 ("")、"(" + A + ")" 或 A + B，其中 A 和 B 都是有效的括号字符串，+ 代表字符串的连接。例如，""，"()"，"(())()" 和 "(()(()))" 都是有效的括号字符串。
如果有效字符串 S 非空，且不存在将其拆分为 S = A+B 的方法，我们称其为原语（primitive），其中 A 和 B 都是非空有效括号字符串。
给出一个非空有效字符串 S，考虑将其进行原语化分解，使得：S = P_1 + P_2 + ... + P_k，其中 P_i 是有效括号字符串原语。
对 S 进行原语化分解，删除分解中每个原语字符串的最外层括号，返回 S 。
示例 :
示例 1：
输入："(()())(())"
输出："()()()"
解释：
输入字符串为 "(()())(())"，原语化分解得到 "(()())" + "(())"，
删除每个部分中的最外层括号后得到 "()()" + "()" = "()()()"。
示例 2：
输入："(()())(())(()(()))"
输出："()()()()(())"
解释：
输入字符串为 "(()())(())(()(()))"，原语化分解得到 "(()())" + "(())" + "(()(()))"，
删除每隔部分中的最外层括号后得到 "()()" + "()" + "()(())" = "()()()()(())"。
示例 3：
输入："()()"
输出：""
解释：
输入字符串为 "()()"，原语化分解得到 "()" + "()"，
删除每个部分中的最外层括号后得到 "" + "" = ""。
提示：
S.length <= 10000
S[i] 为 "(" 或 ")"
S 是一个有效括号字符串
思路:
由于 S一定是一个递归匹配的括号字符串, 如果给左括号赋值 +1, 右括号赋值 -1, 对每一个匹配的原语最后的和一定是 0
那么最外层的括号一定满足: 左括号, count == 0; 右括号, count == 1
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
string removeOuterParentheses(string S)
{
string result = "";
int count = 0;
for(auto c : S)
{
if (c == ')') --count;
if (count > 0) result += c;
if (c == '(') ++count;
}
return result;
}
};
Java:
class Solution {
public String removeOuterParentheses(String S) {
StringBuilder result = new StringBuilder();
int count = 0;
for(char c : S.toCharArray()) {
if (c == ')') --count;
if (count > 0) result.append(c);
if (c == '(') ++count;
}
return result.toString();
}
}
Python:
class Solution:
def removeOuterParentheses(self, S: str) -> str:
count, result = 0, ""
for s in S:
if s == '(':
count += 1
if count == 1:
continue
else:
count -= 1
if not count:
continue
result += s
return result