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给定表达式exp的长度为n，由一些方括号组成。任务是在解析表达式时打印括号编号。

例子 ：

Input : (a+(b*c))+(d/e)
Output : 1 2 2 1 3 3
The highlighted brackets in the given expression
(a+(b*c))+(d/e) has been assigned the numbers as:
1 2 2 1 3 3.

Input : ((())(()))
Output : 1 2 3 3 2 4 5 5 4 1 

资料来源： Flipkart采访经历|设置49。

方法：

1. 定义一个变量left_bnum = 1。
2. 创建一个堆栈right_bnum 。
3. 现在，对于i = 0到n-1。
   1. 如果EXP [I] == ‘(’，然后由1打印到堆栈right_bnum最后增量left_bnum left_bnum，推left_bnum。
   2. 否则，如果exp [i] ==’)’，则打印堆栈的顶部元素right_bnum ，然后从堆栈中弹出顶部元素。

C++

// C++ implementation to print the bracket number
#include 
  
using namespace std;
  
// function to print the bracket number
void printBracketNumber(string exp, int n)
{
    // used to print the bracket number
    // for the left bracket
    int left_bnum = 1;
      
    // used to obtain the bracket number 
    // for the right bracket
    stack right_bnum;
      
    // traverse the given expression 'exp'
    for (int i = 0; i < n; i++) {
          
        // if current character is a left bracket
        if (exp[i] == '(') {
            // print 'left_bnum',
            cout << left_bnum << " ";
              
            // push 'left_bum' on to the stack 'right_bnum'
            right_bnum.push(left_bnum);
              
            // increment 'left_bnum' by 1
            left_bnum++;
        }
          
        // else if current character is a right bracket
        else if(exp[i] == ')') {
  
            // print the top element of stack 'right_bnum'
            // it will be the right bracket number
            cout << right_bnum.top() << " ";
              
            // pop the top element from the stack
            right_bnum.pop();
        }
    }
}
  
// Driver program to test above
int main()
{
    string exp = "(a+(b*c))+(d/e)";
    int n = exp.size();
      
    printBracketNumber(exp, n);
      
    return 0;
}

Java

// Java implementation to 
// print the bracket number
import java.io.*;
import java.util.*;
  
class GFG
{
    // function to print 
    // the bracket number
    static void printBracketNumber(String exp, 
                                   int n)
    {
        // used to print the 
        // bracket number for 
        // the left bracket
        int left_bnum = 1;
          
        // used to obtain the 
        // bracket number for
        // the right bracket
        Stack right_bnum =
                   new Stack();
          
        // traverse the given
        // expression 'exp'
        for (int i = 0; i < n; i++) 
        {
              
            // if current character 
            // is a left bracket
            if (exp.charAt(i) == '(') 
            {
                  
                // print 'left_bnum',
                System.out.print(
                       left_bnum + " ");
                  
                // push 'left_bum' on to
                // the stack 'right_bnum'
                right_bnum.push(left_bnum);
                  
                // increment 'left_bnum' by 1
                left_bnum++;
            }
              
            // else if current character 
            // is a right bracket
            else if(exp.charAt(i) == ')')
            {
      
                // print the top element 
                // of stack 'right_bnum'
                // it will be the right 
                // bracket number
                System.out.print(
                       right_bnum.peek() + " ");
                  
                // pop the top element 
                // from the stack
                right_bnum.pop();
            }
        }
    }
      
    // Driver Code
    public static void main(String args[])
    {
        String exp = "(a+(b*c))+(d/e)";
        int n = exp.length();
          
        printBracketNumber(exp, n);
    }
}
  
// This code is contributed 
// by Manish Shaw(manishshaw1)

Python3

# Python3 implementation to print the bracket number
  
# function to print the bracket number
def printBracketNumber(exp, n):
  
    # used to print the bracket number
    # for the left bracket
    left_bnum = 1
  
    # used to obtain the bracket number
    # for the right bracket
    right_bnum = list()
  
    # traverse the given expression 'exp'
    for i in range(n):
  
        # if current character is a left bracket
        if exp[i] == '(':
  
            # print 'left_bnum',
            print(left_bnum, end = " ")
  
            # push 'left_bum' on to the stack 'right_bnum'
            right_bnum.append(left_bnum)
  
            # increment 'left_bnum' by 1
            left_bnum += 1
  
        # else if current character is a right bracket
        elif exp[i] == ')':
  
            # print the top element of stack 'right_bnum'
            # it will be the right bracket number
            print(right_bnum[-1], end = " ")
  
            # pop the top element from the stack
            right_bnum.pop()
  
# Driver Code
if __name__ == "__main__":
    exp = "(a+(b*c))+(d/e)"
    n = len(exp)
  
    printBracketNumber(exp, n)
  
# This code is contributed by
# sanjeev2552

C#

// C# implementation to 
// print the bracket number
using System;
using System.Collections.Generic;
  
class GFG
{
    // function to print 
    // the bracket number
    static void printBracketNumber(string exp, 
                                   int n)
    {
        // used to print the bracket 
        // number for the left bracket
        int left_bnum = 1;
          
        // used to obtain the bracket  
        // number for the right bracket
        Stack right_bnum = new Stack();
          
        // traverse the given
        // expression 'exp'
        for (int i = 0; i < n; i++) 
        {
              
            // if current character 
            // is a left bracket
            if (exp[i] == '(') 
            {
                  
                // print 'left_bnum',
                Console.Write(left_bnum + " ");
                  
                // Push 'left_bum' on to
                // the stack 'right_bnum'
                right_bnum.Push(left_bnum);
                  
                // increment 'left_bnum' by 1
                left_bnum++;
            }
              
            // else if current character 
            // is a right bracket
            else if(exp[i] == ')')
            {
      
                // print the top element 
                // of stack 'right_bnum'
                // it will be the right 
                // bracket number
                Console.Write(right_bnum.Peek() + " ");
                  
                // Pop the top element 
                // from the stack
                right_bnum.Pop();
            }
        }
    }
      
    // Driver Code
    static void Main()
    {
        string exp = "(a+(b*c))+(d/e)";
        int n = exp.Length;
          
        printBracketNumber(exp, n);
    }
}
  
// This code is contributed 
// by Manish Shaw(manishshaw1)

输出：

1 2 2 1 3 3

时间复杂度： O(n)。
辅助空间： O(n)。