线性结构——用栈实现波兰表达式求值。

//PolishExpression_H.h

#pragma once
//用栈实现兰波表达式求值
#define OK true
#define ERROR false
#define OVERFLOW -1
#define MAXSIZE 100

typedef float ElemType;
typedef bool Status;

typedef struct Stack {
ElemType *base;
ElemType *top;
int stacksize;
}STACK;

Status InitStack(STACK &S);
Status StackFull(STACK S);
Status Push(STACK &S, ElemType e);
Status StackEmpty(STACK S);
ElemType Pop(STACK &S);
void PolishExpression();

//PolishExpression_achieve

#include "PolishExpression_H.h"
#include<stdio.h>
#include<stdlib.h>

Status InitStack(STACK &S) {
S.base = new ElemType[MAXSIZE];
if (!S.base) {
printf("内存分配失败！\n");
exit(OVERFLOW);
}
S.top = S.base;
return OK;
}

Status StackFull(STACK S) {
if (S.top - S.base == MAXSIZE) {
return OK;
}
else return ERROR;
}

Status Push(STACK &S, ElemType e) {
if (StackFull(S)) {
printf("栈满！\n");
return ERROR;
}
*S.top = e;
S.top++;
return OK;
}

Status StackEmpty(STACK S) {
if (S.base == S.top) {
return OK;
}
else return ERROR;
}

ElemType Pop(STACK &S) {
if (StackEmpty(S)) {
return ERROR;
}
ElemType e;
S.top--;
e = *S.top;
return e;
}
//逆波兰表达式(即后缀表达式)求值规则如下：设立运算数栈OPND, 对表达式从左到右扫描(读入)，当表达式中扫描到数时，压入OPND栈。
//当扫描到运算符时，从OPND退出两个数，进行相应运算，结果再压入OPND栈。这个过程一直进行到读出表达式结束符$，这时OPND栈中只有一个数，就是结果。
//以$符作为输入结束，操作数之间用空格分隔,
//操作符只可能有 + 、 - 、*、 / 四种运算。例如：234 34 + 2 * $。

void PolishExpression() {
static int i = 0;
STACK OPND;
InitStack(OPND);
static char ch[30];
printf("请输入表达式（以$符作为输入结束，操作数之间用空格分隔）：\n");
scanf("%c", &ch[i]);

float num = 0.00;
float scale = 0.00;
float x1 = 0.00, x2 = 0.00;
while (ch[i] != '$') {
while ((ch[i] >= '0' && ch[i] <= '9') || ch[i] == '.') {  //拼数
if (ch[i] != '.' && ch[i] != ' ') {				//处理整数
num = num * 10 + (ch[i] - '0');
i++;
scanf("%c", &ch[i]);
}
if (ch[i] == '.') {									//处理小数部分。
scale = 10.0;
i++;
scanf("%c", &ch[i]);
while (ch[i] >= '0' && ch[i] <= '9') {
num = num + (ch[i] - '0') / scale;
scale = scale * 10;
i++;
scanf("%c", &ch[i]);
}
}
}
Push(OPND, num);
num = 0.0;//数压入栈，下个数初始化
scale = 10.0;
if (ch[i] == '+') {
Push(OPND, Pop(OPND) + Pop(OPND));
}
if (ch[i] == '-') {
x1 = Pop(OPND); x2 = Pop(OPND); Push(OPND, x2 - x1);
}
if (ch[i] == '*') {
Push(OPND, Pop(OPND)* Pop(OPND));
}
if (ch[i] == '/') {
x1 = Pop(OPND); x2 = Pop(OPND); Push(OPND, x2 / x1);
}
i++;
scanf("%c", &ch[i]);//读入表达式中下一个字符。
}
printf("后缀表达式的值为: %.2f", Pop(OPND));
printf("\n");
}

#include"PolishExpression_H.h"
#include<stdio.h>
int main() {
PolishExpression();

return OK;
}