原文:
给定一个长度为 n 的表达式 exp ，由一些括号组成。任务是在解析表达式时打印括号数字。 例:
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
来源:t2】flipkart 采访体验|第 49 集。
接近 :
- 定义一个变量 left_bnum = 1。
- 创建一个堆栈 right_bnum 。
- 现在，对于 i = 0 到 n-1。
- 如果 exp[i] == '(')，则打印 left_bnum ，将 left_bnum 推到堆栈 right_bnum 上，最后将 left_bnum 增加 1。
- 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 语言(一种计算机语言，尤用于创建网站)
// 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)
python 3
# 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)
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
output:
1 2 2 1 3 3
时间复杂度: o(n)。 辅助空间: o(n)。
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