逆波兰计算器与中缀表达式向后缀表达式的转化实例

#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>
using namespace std;

typedef char ElemType;
typedef enum{TRUE,FALSE} Status;

#define STACK_INIT_SIZE 20
#define STACKINCREMENT  10
#define MAXBUFFER       10

typedef struct
{
ElemType *base;
ElemType *top;
int stackSize;
}sqStack;

void InitStack(sqStack *s)
{
s->base = (ElemType*)malloc(STACK_INIT_SIZE*sizeof(ElemType));
if (!s->base)
{
exit(0);
}
s->top = s->base;
s->stackSize = STACK_INIT_SIZE;
}

Status Push(sqStack *s, ElemType e)
{
if (s->top - s->base >= s->stackSize)
{
s->base = (ElemType*)realloc(s->base, (s->stackSize + STACKINCREMENT)*sizeof(ElemType));
if (!s->base)
{
return FALSE;
}
s->top = s->base + s->stackSize;
s->stackSize += STACKINCREMENT;
}
*(s->top) = e;
s->top++;
return TRUE;
}

Status Pop(sqStack *s, ElemType *e)
{
if (s->top == s->base)
{
return FALSE;
}
*e = *--(s->top);
return TRUE;
}

int StackLen(sqStack s)
{
return (s.top - s.base);
}

//中缀表达式转换成后缀表达式
//规则：从左到右遍历中缀表达式的每个数字和符号，若是数字直接输出，若是
//符号，则判断其与栈顶的优先级，是右括号或优先级低于栈顶符号，则栈顶元素依次出栈并输出
//直到遇到左括号或栈空才将此符号入栈。

Status MidToEnd(sqStack s)
{
char c, e;
scanf("%c", &c);

while (c != '#')
{
while (c >= '0' && c <= '9')
{
//处理两位及以上的数字
printf("%c", c);
scanf("%c", &c);
if (c < '0' || c > '9')
{
//不是数字时
printf(" ");
}
}
if (')' == c)
{
//为')'时需要出栈查找'('
Pop(&s, &e);
while ('(' != e)
{
printf("%c ", e);
Pop(&s, &e);
if (!StackLen(s))
{
//为空则说明没有'('
//printf("出错，缺少'('！\n");
//失败返回
return FALSE;
}
}
}
else if ('+' == c || '-' == c)
{
if (!StackLen(s))
{
//为空直接压栈
Push(&s, c);
}
else
{
//不为空说明栈内还有符号，需要进行相应优先级的比较
do
{
Pop(&s, &e);
if ('(' == e)
{
//若栈顶元素为'('，直接入栈
Push(&s, e);
}
else
{
//否则输出栈顶元素
printf("%c ", e);
}
} while (StackLen(s) && '(' != e);
//当栈不为空且栈顶值不为'('一直输出
Push(&s, c);
//将当前输入符号入栈
}
}
else if ('*' == c || '/' == c || '(' == c)
{
//直接入栈
Push(&s, c);
}
else if ('#' == c)
{
//由于在判断数字时多次输入，故无法保证是否输入'#'，此处判断
//在上面判断也可
break;
}
else
{
//printf("出错，输入格式错误！\n");
//失败返回
return FALSE;
}
scanf("%c", &c);
//继续输入
}
while (StackLen(s))
{
//若数字遍历完，则清空符号栈
Pop(&s, &e);
printf("%c ", e);
}
//成功返回
return TRUE;
}
//逆波兰计算器
void Cal(sqStack s)
{
char c, d, e;
char str[MAXBUFFER];
//缓冲Buffer
int i = 0;
scanf("%c", &c);
while (c != '#')
{
while (isdigit(c) || c == '.')//用于过滤数字
{
str[i++] = c;
str[i] = '\0';
if (i >= MAXBUFFER)
{
printf("出错，输入的单个数字过大！\n");
return;
}
scanf("%c", &c);
if (c == ' ')//数字输入结束
{
d = atoi(str);//str-->int
Push(&s, d);
i = 0;
break;
}
}
switch (c)
{
case '+':
Pop(&s, &d);
Pop(&s, &e);
Push(&s, d+e);
break;
case '-':
Pop(&s, &d);
Pop(&s, &e);
Push(&s, e-d);
break;
case '*':
Pop(&s, &d);
Pop(&s, &e);
Push(&s, d*e);
break;
case '/':
Pop(&s, &d);
Pop(&s, &e);
if (d != 0)
{
Push(&s, e/d);
}
else
{
printf("\n出错，除数为零！\n");
return;
}
break;
default:
break;
}
scanf("%c", &c);
}
Pop(&s, &d);
//得出结果
printf("最终计算结果为：%d\n",d);
}

int main()
{
sqStack s;
InitStack(&s);
printf("请输入中缀表达式，以#为结束标志：");
if (MidToEnd(s) == TRUE)
{
printf("\n上式为逆波兰表达式结果，请输入#结束输入：\n");
Cal(s);
}
else
{
printf("\n出错，输入格式错误！\n");
}

return 0;
}