华为2012校招成都站8:00最后一题，表达式计算源代码

/*表达式计算-华为校招
Author : Milo.Wang
Date : 2012/9/15
*/
#include <iostream>
#include <cstring>
using namespace std;

void Expr(const char *expr, int *ret)
{
char expr1[1024];
char expr2[1024];
strcpy(expr1,expr);
strcpy(expr2,expr);
int opds[1024];
char oprs[1024];
int idxd = 0, idxr = 0;
int opd_idx = -1;
char * p = strtok(expr1,"()*+-");
while(p != NULL)
{
opds[idxd++] = atoi(p);
p = strtok(NULL,"()*+-");
}
int i;
int digit_flag = true;
for(i = 0; expr2[i] != '\0'; ++i)
{
switch(expr2[i])
{
case '(':
case ')':
case '+':
case '-':
case '*':
digit_flag = true;
oprs[idxr++] = expr2[i];
break;
default:
if(digit_flag)
{
digit_flag = false;
oprs[idxr++] = '#';//特殊符号，标示有一个数字
}
}
}
oprs[idxr++] = '\0';

int opd_stack[1024];
char opr_stack[1024];
int opd_top = -1;
int opr_top = -1;
int opdr,opdl;
for(i = 0; i < idxr; ++i)
{
char ch;
switch((ch = oprs[i]))
{
case '*':
case '(':
opr_stack[++opr_top] = ch;
break;
case '+':
case '-':
while(opr_top > -1 && opr_stack[opr_top] == '*')
{
--opr_top;
opdr = opd_stack[opd_top--];
opdl = opd_stack[opd_top--];
opd_stack[++opd_top] = opdl * opdr;
}
opr_stack[++opr_top] = ch;
break;
case ')':
while(opr_top > -1 && opr_stack[opr_top] != '(')
{
opdr = opd_stack[opd_top--];
opdl = opd_stack[opd_top--];
char c;
int result;
switch((c = opr_stack[opr_top--]))
{
case '*':
result = opdl * opdr;
break;
case '+':
result = opdl + opdr;
break;
case '-':
result = opdl - opdr;
break;
}
opd_stack[++opd_top] = result;
}
if(opr_stack[opr_top] == '(')
--opr_top;
break;
case '\0':
while(opr_top > -1)
{
opdr = opd_stack[opd_top--];
opdl = opd_stack[opd_top--];
char c;
int result;
switch((c = opr_stack[opr_top--]))
{
case '*':
result = opdl * opdr;
break;
case '+':
result = opdl + opdr;
break;
case '-':
result = opdl - opdr;
break;
}
opd_stack[++opd_top] = result;
}
break;
case '#':
opd_stack[++opd_top] = opds[++opd_idx];
}
}
*ret = opd_stack[opd_top];
}
int main(int argc, char *argv[])
{
const char *str = "5+6*(5-2)-10";
int ret;
Expr(str,&ret);
cout << str << " = " << ret << endl;
return 0;
}