"C + +" is picking up the flower--infix suffix

For simple arithmetic, postfix expressions can be used to quickly calculate results by using stacks (stack)

================================== I'm a split line ====================================

Definition of suffix:

e.g. 2 + 3, 2 3 +

2 + 3 * 4-2 3 4 * +

Apply stack to calculate suffix expression:

e.g. suffix expression 6 5 2 3 + 8 * + 3 + *

Traversal: 6 push (6) Stack:6

5 push (5) Stack:6 5

2 push (2) Stack:6 5 2

3 Push (3) Stack:6 5 2 3

+ Pop, Pop 2, 3 out of stack, operation ans = 2 + 3; push (ANS) stack:6 5 5

8 Push (8) Stack:6 5 5 8

* Pop, Pop 5, 8 out of the stack, operation ans = 5 * 8; push (ANS) stack:6 5 40

+ Pop, pop 5, 40 out of stack, operation ans = 5 + +; push (ans) stack:6 45

3 Push (3) Stack:6 45 3

+ Pop, pop 45, 3 out of stack, operation ans = 3; push (ans) Stack:6 48

* Pop, pop 6, 48 out of the stack, operation ans = 6 *, push (ans) stack:288

Converts an infix expression to a suffix expression:

Setting: Priority ' (' > ' * ' = '/' > ' + ' = '-')

① reads the character of the input queue, judging whether it is a number or a symbol +-*/()

② If a number, put it in the output queue

③ if ') ', the symbol one by one in the stack is ejected to the output queue until the first ' (', and ' (' not placed in the output queue ') is encountered

④ if the other symbol is +-*/(, the element in the stack pops up until a lower-priority symbol is found or ' ('). ' (' only when encountering ') ' only pops up

Code Implementation :

　

1 //2016-03-162 //infix suffix3 4 //Limitations: Enter a range of legal, digital 0~95#include <iostream>6#include <string>7#include <stack>8 9 using namespacestd;Ten  One intMain () { A     stringinput, output; -stack<Char>operators; -      while(Cin >>input) { the         if(Input = ="0") Break; - output.clear (); -          for(inti =0; I < input.size (); i++) { -             CharCH =Input[i]; +             if(Ch >='0'&& CH <='9') output.push_back (CH); -             Else if(ch = ='+'|| ch = ='-') { +                  while(!operators.empty () && (operators.top () = ='*'|| Operators.top () = ='/'|| Operators.top () = ='-'|| Operators.top () = ='+')) { A Output.push_back (Operators.top ()); at Operators.pop (); -                 } - Operators.push (CH); -             } -             Else if(ch = ='*'|| ch = ='/') { -                  while(!operators.empty () && (operators.top () = ='*'|| Operators.top () = ='/')) { in Output.push_back (Operators.top ()); - Operators.pop (); to                 } + Operators.push (CH); -             } the             Else if(ch = =')') { *                  while(Operators.top ()! ='('&&!Operators.empty ()) { $ Output.push_back (Operators.top ());Panax Notoginseng Operators.pop (); -                 } the Operators.pop (); +             } A             Else if(ch = ='(') { the Operators.push (CH); +             } -         } $          while(!Operators.empty ()) { $ Output.push_back (Operators.top ()); - Operators.pop (); -         } thecout <<"Output:"<< Output <<Endl; -     }Wuyi     return 0; the}

Test :

　　

Limitations: (not strong enough)

① only implements the digital input of the 0~9, if more than two digits are required to make a character judgement (whether the previous character is a number, if so, then the current character and the previous one is the same integer)

② does not do the final result calculation, because it is still a string, if necessary, to do the character to integer conversion judgment.

③ No anomaly detection (assuming all inputs are valid, illegal input will directly result in incorrect output)

some bugs:

No clear () in the ①stack container;

② judgment Condition while (!operators.empty () && (operators.top () = = ' * ' | | operators.top () = = '/') must first be judged by the!operators.empty ()

Otherwise the first judgment (operators.top () = = ' * ' | | operators.top () = = '/') is actually the top () of the read operation, when the stack is empty, the runtime error

"C + +" is picking up the flower--infix suffix