Algorithm title: Arithmetic (infix to postfix conversion, the inverse of the Polish style worth thinking about)

/ * Arithmetic of the string. Give a string that contains the 0~9 number and the +-*\/() operator,-only the minus sign does not represent a negative. Examples are as follows: Input: 1 + 2 * (3-4) *///Haha, see this problem, in fact, it is not difficult, the problem is not to think,//Of course you understand the algorithm, the algorithm used here is inverse Polish style. //If you do not understand the place, you can search the Internet to reverse the Polish style. /* My summary: The computer can not understand the human positive thinking, so in order to meet the computer's thinking, we will do the opposite, the operation symbol is placed behind the operand, to form a suffix expression, but if you can according to the Encyclopedia of the thinking to do, then easy to get the answer. *//*1. Constructs an inverse Polish expression based on an expression. 2. Calculation according to the inverse Polish expression. 3. The result can be easily obtained in the end, I hope you can learn, mutual encouragement. *///start:#include <iostream>#include <string.h>#include <stack>using namespace STD;BOOLIs_op (CharCH) {return(ch = =' + '|| ch = ='-'|| ch = =' * '|| ch = ='/'|| ch = =' ('|| ch = =' ) '|| ch = =' # ');}BOOLIs_com (CharCH1,CharCH2) {//Compare symbol expression is priority.     inti =0;intj =0; for(;"#+-*/()"[i]! = CH1; i++) {} for(;"#+-*/()"[j]! = CH2; J + +) {}BOOLstr[][7] = {1,0,0,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,1,1,0,0,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1};returnSTR[I][J];}intGrial (Char*&AMP;STR) {//Check if the brace match is ok first.     intFlags =0;Char*FP = str; while(*FP! =' + ')    {if(*FP = =' (') flags++;if(*FP = =' ) ') flags--;    fp++; }if(Flags! =0)return-1;//Start constructing Polish expressions.     strcat(STR,"#");Char*p = str;Char*q = str; Stack<char>OP; Op.push (' # '); Stack<string>NM;intCount =0;stringS while(*p! =' + ')    {if(Is_op (*P)) {if(*p = =' ) ')            { while(Op.top ()! =' (') {s = Op.top ();                    Nm.push (s);                Op.pop (); }if(Op.top () = =' (') {Op.pop (); p++;Continue; }            }if(Is_com (Op.top (), *p) && op.top ()! =' (')            { while(Is_com (Op.top (), *p))                    {s = op.top ();                    Nm.push (s); Op.pop ();if(*p = =' # '&& op.top () = =' # ') Break;            } op.push (*p); }Else{Op.push (*p);        } p++; }Else{ while(! Is_op (*p)) {count = Count *Ten+ *p-' 0 '; p++;if(*p = =' + '|| *p = =' # ') Break; }Charbuff[255];memset(Buff,0,sizeof(buff)); Itoa (count, Buff,Ten);            s = buff;            Nm.push (s); Count =0; }    } Stack<string>Numst; while(Nm.empty () = =false) {Numst.push (Nm.top ());//reverse order. Nm.pop (); }//So far the inverse Polish style has been constructed, let's calculate it.      Stack<int>Temp while(Numst.empty () = =false)    {if(Numst.top () = ="+"|| Numst.top () = ="-"|| Numst.top () = ="*"|| Numst.top () = ="/")        {intA = Temp.top (); Temp.pop ();intb = Temp.top (); Temp.pop ();if(Numst.top () = ="+") {Temp.push (b + a); }Else if(Numst.top () = ="-") {Temp.push (b-a); }Else if(Numst.top () = ="*") {Temp.push (b*a); }Else if(Numst.top () = ="/") {Temp.push (b/a);        } numst.pop (); }Else{Temp.push (Atoi (Numst.top (). C_STR ()));        Numst.pop (); }    }returnTemp.top ();}intMain () {;Char*s =New Char[ -];memset(S,0,sizeof(s));strcpy(S,"(1+2) *3-(1+2)/4+10");cout<< grial (s) << Endl;strcpy(S,"1+2+4+ (1*2)/3+4");cout<< grial (s) << Endl;strcpy(S,"+1+2-1*2" (4/2));cout<< grial (s) << Endl;strcpy(S,(1+ (2+3) +2-0));//I am also concerned that my double brackets are not taken into account,    //Didn't think of a success.     cout<< grial (s) << Endl;return 0;}

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Algorithm title: Arithmetic (infix to postfix conversion, the inverse of the Polish style worth thinking about)