添加缺少左括号入式 [英] Add missing left parentheses into equation

问题描述

我是pretty的停留在找到一个给定的问题得到妥善解决，一直在寻找一些想法在互联网上。没有能找到任何

I'm pretty stuck in finding a proper solution for a given problem, been looking for some ideas over the internet. Wasn't be able to find any.

问题是：写一个程序，它从标准输入的前pression没有留下括号并打印相当于缀EX pression插入了括号

Problem is: writing a program that takes from the standard input an expression without left parentheses and prints the equivalent infix expression with the parentheses inserted.

由于EX pression： 1 + 2）* 3 - 4）* 5 - 6）））
输出：（（1 + 2）*（（3 - 4）*（5 - 6）））

Given expression: 1 + 2) * 3 - 4)* 5 - 6)))
Output: ((1 + 2) * ((3 - 4) * (5 - 6)))

有什么可以解决这个问题最好的办法？

What can be the best approach to solve this problem?

推荐答案

我觉得，我们的目标是假设你只插EX pressions，而不是孤独的数字。

I think that the goal is assuming that you only parenthesize expressions, not lone numbers.

所以，你要抓住每个令牌折腾他们在一个堆栈

So you'll want to grab each token and toss them on a stack

2
+
1

抓住下一个标记，这是） 现在就堆栈的前三名和夹心它的括号（1 + 2）之间，将其放回栈为一体EX pression上。

grab the next token, which is ) now take the top three of the stack and sandwich it between those parens ( 1 + 2 ), put it back on the stack as one expression.

接下来的推栈看起来像这样

next push the stack looks like this

4
-
3
*
(1 + 2)

拉​​出前三名，并把它放回栈括号上     （3-4）     *     （1 + 2）

pull out the top three and put it back on the stack parenthesized (3-4) * (1+2)

和再

6
-
5
*
(3-4)
*
(1+2)

我们再创括号并再次抢前3名从堆栈，插推背

we hit another paren and grab the top 3 from the stack again, parenthesize and push back

(5-6)
*
(3-4)
*
(1+2)

我们去抢另一个的括号，抢前3名从堆栈中再次，插推背

We grab another paren, grab the top 3 from the stack again, parenthesize and push back

((3-4)*(5-6))
*
(1+2)

和再...

((1 + 2) * ((3 - 4) * (5 - 6)))

没有更多的投入，所以这是我们的答案

no more input, so this is our answer

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