栈应用 - 后缀表达式的计算

有关栈API详情参看我的另一篇博文： 栈的链式存储 - API实现

遍历后缀表达式中的数字和符号

对于数字：进栈

对于符号：

从栈中弹出右操作数

从栈中弹出左操作数

根据符号进行运算

将运算结果压入栈中

遍历结束：栈中的唯一数字为计算结果

#include <stdio.h>
#include "LinkStack.h"

int isNumber3(char c)
{
return ('0' <= c) && (c <= '9');
}

int isOperator3(char c)
{
return (c == '+') || (c == '-') || (c == '*') || (c == '/');
}

int value(char c)
{
return (c - '0');
}

int express(int left, int right, char op)
{
int ret = 0;

switch (op)
{
case '+':
ret = left + right;
break;
case '-':
ret = left - right;
break;
case '*':
ret = left * right;
break;
case '/':
ret = left / right;
break;
default:
break;
}

return ret;
}

int compute(const char* exp)
{
LinkStack* stack = LinkStack_Create();
int ret = 0;
int i = 0;

while (exp[i] != '\0')
{
if (isNumber3(exp[i]))
{
LinkStack_Push(stack, (void*)value(exp[i]));
}
else if (isOperator3(exp[i]))
{
int right = (int)LinkStack_Pop(stack);
int left = (int)LinkStack_Pop(stack);
int result = express(left, right, exp[i]);

LinkStack_Push(stack, (void*)result);
}
else
{
printf("Invalid expression!");
break;
}

i++;
}

if ((LinkStack_Size(stack) == 1) && (exp[i] == '\0'))
{
ret = (int)LinkStack_Pop(stack);
}
else
{
printf("Invalid expression!");
}

LinkStack_Destroy(stack);

return ret;
}

int main()
{
printf("8 + (3 - 1) * 5  = %d\n", compute("831-5*+"));

return 0;
}

工程文件详情：Github