Data structure Practice-suffix expression (stack)

This article is for the data structure Basic Series Network course (3): Stack and queue implementation project.

"Project-suffix expression"

The algorithm of converting an infix expression to the corresponding suffix expression is realized by using the basic operation of the stack in Sqstack.h. For example, input (56-20)/(4+2), output suffix expression:: 56#20#-4#2#+/requires the number to be appended with #.

Description of the reference solution:

Principle see the application of video stack 1-expression evaluation, you can also refer to the expression evaluation of the source code
Based on the stack structure, the algorithm steps to convert infix expressions to Postfix expressions are:

Initializes the operator stack op;' = 'into the stack;ExpRead character ch; while(ch!=' + '){if(Ch is not an operator) all subsequent digits are stored in the Postexp, followed by a character' # 'Flag value string End;Else        Switch(Precede (OP stack top operator, CH)) { Case ' < '://Stack top operator priority LowThe CH into the stack; FromExpRead the next character ch; Break; Case ' = '://Only the top of the stack operator is ' (', ch ') ' Casefallback stack; FromExpRead the next character ch; Break; Case ' > '://stack top operator should be executed first, so the stack and put into the PostexpStack up operator and store it in Postexp; Break; }}

The implementation code is as follows:

header file sqstack.h see [Sequential Stack Algorithm library], using a chain stack is also possible.

#include <stdio.h>#include <stdlib.h>#include "sqstack.h"#define MAXOP 7struct  //Set operator precedence{CharCh//Operator    intpri//Priority}lpri[]= {{' = ',0},{' (',1},{' * ',5},{'/',5},{' + ',3},{'-',3},{' ) ',6}},rpri[]= {{' = ',0},{' (',6},{' * ',4},{'/',4},{' + ',2},{'-',2},{' ) ',1}};intLeftpri (CharOp//To prioritize left operator op{intI for(i=0; i<maxop; i++)if(LPRI[I].CH==OP)returnLpri[i].pri;}intRightpri (CharOp//The priority of the right operator op{intI for(i=0; i<maxop; i++)if(RPRI[I].CH==OP)returnRpri[i].pri;}BOOLInop (CharCh//Determine if CH is an operator{if(ch==' ('|| ch==' ) '|| ch==' + '|| ch=='-'|| ch==' * '|| ch=='/')return true;Else        return false;}intPrecede (CharOP1,CharOP2)comparison results for//OP1 and OP2 operator Precedence{if(Leftpri (OP1) ==rightpri (OP2))return 0;Else if(Leftpri (OP1) <rightpri (OP2))return-1;Else        return 1;}voidTransChar*Exp,CharPostexp[])//Convert an arithmetic expression exp to a suffix expression postexp{Sqstack *opstack;//define operator Stacks    intI=0;//i as the subscript of PostexpElemtype ch; Initstack (Opstack);//Use initialization stack operation to allocate space for stack, make sure to doPush (Opstack,' = '); while(*Exp!=' + ')//exp loop When expression is not finished scanning{if(! INOP (*Exp))//Case of numeric characters{ while(*Exp>=' 0 '&& *Exp<=' 9 ')//judged as digital{postexp[i++]=*Exp;Exp++; } postexp[i++]=' # ';//Use # to mark the end of a numeric string}Else    //For the case of the operator{GetTop (opstack, ch);//Get top of stack operator            Switch(Precede (CH, *Exp))            { Case-1://stack top operator with low priority: in-StackPush (Opstack, *Exp);Exp++;//Continue to scan other characters                 Break; Case 0://Only brackets to meet this situationPop (Opstack, ch);//Will (back up the stack                Exp++;//Continue to scan other characters                 Break; Case 1://rewind stack and output to PostexpPostexp[i++]=ch; Pop (Opstack, ch); Break; }        }    }//while (*exp!= ')Pop (Opstack, ch); while(ch!=' = ')//At this time exp scan complete, back to ' = ' so far{postexp[i++]=ch;    Pop (Opstack, ch); } postexp[i]=' + ';//Add end-of-identity to postexp expressionDestroystack (opstack);}intMain () {Char Exp[]="(56-20)/(4+2)";//exp can be changed to keyboard input    Charpostexp[ $]; TransExp, postexp);printf("infix expression:%s\n",Exp);printf("suffix expression:%s\n", postexp);return 0;}

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Data structure Practice-suffix expression (stack)