从包含 + 和 –运算符的代数字符串中删除括号

简化给定的代数字符串，“+”、“-”运算符和括号。输出不带括号的简化字符串。
例子：

Input : "(a-(b+c)+d)"
Output : "a-b-c+d"

Input : "a-(b-c-(d+e))-f"
Output : "a-b+c+d+e-f" 

这个想法是在括号开始之前检查运算符，即在字符'（'之前。如果运算符是 - ，我们需要切换括号内的所有运算符。使用一个堆栈，它只存储两个整数 0 和 1 来指示是否切换或不切换。
我们对输入字符串的每个字符进行迭代。最初将 0 压入堆栈。每当字符是运算符（'+' 或 '-'）时，检查堆栈顶部。如果栈顶为 0，则在结果字符串中附加相同的运算符。如果栈顶为 1，则在结果字符串中附加另一个运算符（如果 '+' 附加 '-'）。

C++

// C++ program to simplify algebraic string
#include 
using namespace std;
 
// Function to simplify the string
char* simplify(string str)
{
    int len = str.length();
 
    // resultant string of max length equal
    // to length of input string
    char* res = new char(len);
    int index = 0, i = 0;
 
    // create empty stack
    stack s;
    s.push(0);
 
    while (i < len) {
          // Don't do any operation
        if(str[i] == '(' && i == 0) {
            i++;
              continue;
        }
       
        if (str[i] == '+') {
 
            // If top is 1, flip the operator
            if (s.top() == 1)
                res[index++] = '-';
 
            // If top is 0, append the same operator
            if (s.top() == 0)
                res[index++] = '+';
 
        } else if (str[i] == '-') {
            if (s.top() == 1) {
                  if(res[index-1] == '+' || res[index-1] == '-') res[index-1] = '+'; // Overriding previous sign
                else res[index++] = '+';
            }
            else if (s.top() == 0) {
                  if(res[index-1] == '+' || res[index-1] == '-') res[index-1] = '-'; // Overriding previous sign
                else res[index++] = '-';
            }
        } else if (str[i] == '(' && i > 0) {
            if (str[i - 1] == '-') {
 
                // x is opposite to the top of stack
                int x = (s.top() == 1) ? 0 : 1;
                s.push(x);
            }
 
            // push value equal to top of the stack
            else if (str[i - 1] == '+')
                s.push(s.top());
        }
 
        // If closing parentheses pop the stack once
        else if (str[i] == ')')
            s.pop();
 
        // copy the character to the result
        else
            res[index++] = str[i];
        i++;
    }
    return res;
}
 
// Driver program
int main()
{
    string s1 = "(a-(b+c)+d)";
    string s2 = "a-(b-c-(d+e))-f";
    cout << simplify(s1) << endl;
    cout << simplify(s2) << endl;
    return 0;
}

Java

// Java program to simplify algebraic string
import java.util.*;
class GfG {
 
// Function to simplify the string
static String simplify(String str)
{
    int len = str.length();
 
    // resultant string of max length equal
    // to length of input string
    char res[] = new char[len];
    int index = 0, i = 0;
 
    // create empty stack
    Stack s = new Stack ();
    s.push(0);
 
    while (i < len) {
          // Don't do any operation
        if(str.charAt(i) == '(' && i == 0) {
            i++;
              continue;
        }
       
        if (str.charAt(i) == '+') {
 
            // If top is 1, flip the operator
            if (s.peek() == 1)
                res[index++] = '-';
 
            // If top is 0, append the same operator
            if (s.peek() == 0)
                res[index++] = '+';
 
        } else if (str.charAt(i) == '-') {
            if (s.peek() == 1)
                res[index++] = '+';
            else if (s.peek() == 0)
                res[index++] = '-';
        } else if (str.charAt(i) == '(' && i > 0) {
            if (str.charAt(i - 1) == '-') {
 
                // x is opposite to the top of stack
                int x = (s.peek() == 1) ? 0 : 1;
                s.push(x);
            }
 
            // push value equal to top of the stack
            else if (str.charAt(i - 1) == '+')
                s.push(s.peek());
        }
 
        // If closing parentheses pop the stack once
        else if (str.charAt(i) == ')')
            s.pop();
               
        else if(str.charAt(i) == '(' && i == 0)
              i = 0;
        // copy the character to the result
          else
            res[index++] = str.charAt(i);
        i++;
    }
    return new String(res);
}
 
// Driver program
public static void main(String[] args)
{
    String s1 = "(a-(b+c)+d)";
    String s2 = "a-(b-c-(d+e))-f";
    System.out.println(simplify(s1));
    System.out.println(simplify(s2));
}
}

Python3

# Python3 program to simplify algebraic String
 
# Function to simplify the String
 
 
def simplify(Str):
    Len = len(Str)
 
    # resultant String of max Length
    # equal to Length of input String
    res = [None] * Len
    index = 0
    i = 0
 
    # create empty stack
    s = []
    s.append(0)
 
    while (i < Len):
          if (Str[i] == '(' and i == 0):
                i += 1
                continue
 
        if (Str[i] == '+'):
 
            # If top is 1, flip the operator
            if (s[-1] == 1):
                res[index] = '-'
                index += 1
 
            # If top is 0, append the
            # same operator
            if (s[-1] == 0):
                res[index] = '+'
                index += 1
 
        else if (Str[i] == '-'):
            if (s[-1] == 1):
                res[index] = '+'
                index += 1
            else if (s[-1] == 0):
                res[index] = '-'
                index += 1
        else if (Str[i] == '(' and i > 0):
            if (Str[i - 1] == '-'):
 
                # x is opposite to the top of stack
                x = 0 if (s[-1] == 1) else 1
                s.append(x)
 
            # append value equal to top of the stack
            else if (Str[i - 1] == '+'):
                s.append(s[-1])
 
        # If closing parentheses pop
        # the stack once
        else if (Str[i] == ')'):
            s.pop()
 
        # copy the character to the result
        else:
            res[index] = Str[i]
            index += 1
        i += 1
    return res
 
# Driver Code
if __name__ == '__main__':
 
    s1 = "(a-(b+c)+d)"
    s2 = "a-(b-c-(d+e))-f"
    r1 = simplify(s1)
    for i in r1:
        if i != None:
            print(i, end = " ")
        else:
            break
    print()
    r2 = simplify(s2)
    for i in r2:
        if i != None:
            print(i, end = " ")
        else:
            break
 
# This code is contributed by PranchalK

C#

// C# program to simplify algebraic string
using System;
using System.Collections.Generic;
 
class GfG
{
 
// Function to simplify the string
static String simplify(String str)
{
    int len = str.Length;
 
    // resultant string of max length equal
    // to length of input string
    char []res = new char[len];
    int index = 0, i = 0;
 
    // create empty stack
    Stack s = new Stack ();
    s.Push(0);
 
    while (i < len)
    {
          // Don't do any operation
        if(str[i] == '(' && i == 0)
        {
            i++;
              continue;
        }
       
        if (str[i] == '+')
        {
 
            // If top is 1, flip the operator
            if (s.Peek() == 1)
                res[index++] = '-';
 
            // If top is 0, append the same operator
            if (s.Peek() == 0)
                res[index++] = '+';
 
        }
        else if (str[i] == '-')
        {
            if (s.Peek() == 1)
                res[index++] = '+';
            else if (s.Peek() == 0)
                res[index++] = '-';
        }
        else if (str[i] == '(' && i > 0)
        {
            if (str[i - 1] == '-')
            {
 
                // x is opposite to the top of stack
                int x = (s.Peek() == 1) ? 0 : 1;
                s.Push(x);
            }
 
            // push value equal to top of the stack
            else if (str[i - 1] == '+')
                s.Push(s.Peek());
        }
 
        // If closing parentheses pop the stack once
        else if (str[i]== ')')
            s.Pop();
       
        // copy the character to the result
        else
            res[index++] = str[i];
        i++;
    }
    return new String(res);
}
 
// Driver code
public static void Main(String[] args)
{
    String s1 = "(a-(b+c)+d)";
    String s2 = "a-(b-c-(d+e))-f";
    Console.WriteLine(simplify(s1));
    Console.WriteLine(simplify(s2));
}
}
 
// This code has been contributed by 29AjayKumar

Javascript

// Javascript program to simplify algebraic string
 
// Function to simplify the string
function simplify(str)
{
    let len = str.length;
   
    // resultant string of max length equal
    // to length of input string
    let res = new Array(len);
    let index = 0, i = 0;
   
    // create empty stack
    let s = [];
    s.push(0);
   
    while (i < len) {
        // Don't do any operation
        if(str[i] == '(' && i == 0) {
            i++;
              continue;
        }
             
        if (str[i] == '+') {
   
            // If top is 1, flip the operator
            if (s[s.length-1] == 1)
                res[index++] = '-';
   
            // If top is 0, append the same operator
            if (s[s.length-1] == 0)
                res[index++] = '+';
   
        } else if (str[i] == '-') {
            if (s[s.length-1] == 1)
                res[index++] = '+';
            else if (s[s.length-1] == 0)
                res[index++] = '-';
        } else if (str[i] == '(' && i > 0) {
            if (str[i - 1] == '-') {
   
                // x is opposite to the top of stack
                let x = (s[s.length-1] == 1) ? 0 : 1;
                s.push(x);
            }
   
            // push value equal to top of the stack
            else if (str[i - 1] == '+')
                s.push(s[s.length-1]);
        }
   
        // If closing parentheses pop the stack once
        else if (str[i] == ')')
            s.pop();
             
        // copy the character to the result
        else
            res[index++] = str[i];
        i++;
    }
    return (res).join("");
}
 
// Driver program
let s1 = "(a-(b+c)+d)";
let s2 = "a-(b-c-(d+e))-f";
document.write(simplify(s1)+"<br>");
document.write(simplify(s2)+"<br>");
 
 
// This code is contributed by rag2127

输出

a-b-c+d
a-b+c+d+e-f