有关栈API详情參看我的还有一篇博文： 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
依据符号进行运算
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;
}