通过从每个原始子字符串删除最外层的括号来减少字符串

给定有效括号的字符串S“(”和“)”中，任务是打印通过除去选自S每原始子串的最外括号中获得的字符串。

A valid parentheses substring S is primitive if it is non-empty, and cannot be split into two or more non-empty substrings which are also a valid parentheses.

例子：

Input: S = “(()())(())()” 
Output: ()()() 
Explanation: The input string is “(()())(())()” can be decomposed into primitive susbstrings “(()())” + “(())”+”()”. After removing outermost parentheses of each priiimitive substrings, the string obtained is “()()” + “()” = “()()()”

Input: S = “((()())(())(()(())))” 
Output: ()()()()(())

处理方法：按照以下步骤解决问题：

1. 初始化一个变量count来存储左括号的数量，即‘(‘ 。
2. 如果count大于0 ，则将每个‘(‘添加到结果中，即在遇到原始子字符串的第一个‘(‘之后添加所有‘(‘ 。
3. 如果count大于0 ，则将每个‘)’添加到结果中，即在遇到原始子字符串的最后一个‘)’之前添加所有‘)’ 。
4. 最后，打印得到的结果字符串。

以下是上述方法的实现-

C++

// C++ program to implement the
// above approach
#include 
using namespace std;
 
// Function to remove the outermost
// parentheses of every primitive
// substring from the given string
string removeOuterParentheses(string S)
{
 
    // Stores the resultant string
    string res;
 
    // Stores the count of
    // opened parentheses
    int count = 0;
 
    // Traverse the string
    for (char c : S) {
 
        // If opening parenthesis is
        // encountered and their
        // count exceeds 0
        if (c == '(' && count++ > 0)
 
            // Include the character
            res += c;
 
        // If closing parenthesis is
        // encountered and their
        // count is less than count
        // of opening parentheses
        if (c == ')' && count-- > 1)
 
            // Include the character
            res += c;
    }
 
    // Return the resultant string
    return res;
}
 
// Driver Code
int main()
{
    string S = "(()())(())()";
    cout << removeOuterParentheses(S);
}

Java

// Java program to implement the
// above approach
import java.io.*;
class GFG{
   
// Function to remove the outermost
// parentheses of every primitive
// substring from the given string
static String removeOuterParentheses(String S)
{
  // Stores the resultant
  // string
  String res = "";
 
  // Stores the count of
  // opened parentheses
  int count = 0;
 
  // Traverse the string
  for (int c = 0;
           c < S.length(); c++)
  {
    // If opening parenthesis is
    // encountered and their
    // count exceeds 0
    if (S.charAt(c) == '(' &&
        count++ > 0)
 
      // Include the character
      res += S.charAt(c);
 
    // If closing parenthesis is
    // encountered and their
    // count is less than count
    // of opening parentheses
    if (S.charAt(c) == ')' &&
        count-- > 1)
 
      // Include the character
      res += S.charAt(c);
  }
 
  // Return the resultant string
  return res;
}
 
// Driver Code
public static void main(String[] args)
{
  String S = "(()())(())()";
  System.out.print(removeOuterParentheses(S));
}
}
 
// This code is contributed by Chitranayal

Python3

# Python3 program to implement the
# above approach
 
# Function to remove the outermost
# parentheses of every primitive
# substring from the given string
def removeOuterParentheses(S):
     
    # Stores the resultant string
    res = ""
 
    # Stores the count of
    # opened parentheses
    count = 0
 
    # Traverse the string
    for c in S:
         
        # If opening parenthesis is
        # encountered and their
        # count exceeds 0
        if (c == '(' and count > 0):
 
            # Include the character
            res += c
 
        # If closing parenthesis is
        # encountered and their
        # count is less than count
        # of opening parentheses
        if (c == '('):
            count += 1
        if (c == ')' and count > 1):
 
            # Include the character
            res += c
             
        if (c == ')'):
          count -= 1
     
    # Return the resultant string
    return res
 
# Driver Code
if __name__ == '__main__':
     
    S = "(()())(())()"
     
    print(removeOuterParentheses(S))
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program to implement
// the above approach 
using System;
 
class GFG{
    
// Function to remove the outermost
// parentheses of every primitive
// substring from the given string
static string removeOuterParentheses(string S)
{
   
  // Stores the resultant
  // string
  string res = "";
  
  // Stores the count of
  // opened parentheses
  int count = 0;
  
  // Traverse the string
  for(int c = 0; c < S.Length; c++)
  {
     
    // If opening parenthesis is
    // encountered and their
    // count exceeds 0
    if (S == '(' &&
        count++ > 0)
  
      // Include the character
      res += S;
  
    // If closing parenthesis is
    // encountered and their
    // count is less than count
    // of opening parentheses
    if (S == ')' &&
        count-- > 1)
  
      // Include the character
      res += S;
  }
  
  // Return the resultant string
  return res;
}
  
// Driver Code
public static void Main()
{
  string S = "(()())(())()";
   
  Console.Write(removeOuterParentheses(S));
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
// Javascript program to implement the
// above approach
 
// Function to remove the outermost
// parentheses of every primitive
// substring from the given string
function removeOuterParentheses(S)
{
 
// Stores the resultant
// string
let res = "";
 
// Stores the count of
// opened parentheses
let count = 0;
 
// Traverse the string
for (let c = 0;
        c < S.length; c++)
{
    // If opening parenthesis is
    // encountered and their
    // count exceeds 0
    if (S.charAt(c) == '(' &&
        count++ > 0)
 
    // Include the character
    res += S.charAt(c);
 
    // If closing parenthesis is
    // encountered and their
    // count is less than count
    // of opening parentheses
    if (S.charAt(c) == ')' &&
        count-- > 1)
 
    // Include the character
    res += S.charAt(c);
}
 
// Return the resultant string
return res;
}
 
// Driver Code
 
let S = "(()())(())()";
document.write(removeOuterParentheses(S));
 
// This code is contributed by jana_sayantan.

输出：

()()()

时间复杂度： O(N)
辅助空间： O(N)

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