hdu1237简单计算器（栈的简单运用）

题目非常简单，也是栈的经典应用。这道题是没有括号的，还有一种四则运算是有括号的。

我们把平时所用的标准四则运算表达式，即“9+(3-1)*3+10/2”叫做中缀表达式。因为所有的运算符号都在两数字的中间，现在我们的问题就是中缀到后缀的转化。

中缀表达式“9+(3-1)3+10/2”转化为后缀表达式“9 3 1-3+ 10 2/+”

后缀表达式方便计算机计算，所以我们要把中缀表达式转换成后缀表达式。我可以用两个栈，一个是保存数字的数字栈，一个是保存符号的符号栈。

这篇文章很好的表示转换的过程：

http://www.nowamagic.net/librarys/veda/detail/2307

除此之外，这个问题还有个关键是对两个符号的比较：

我把符号的比较分为两个情况，一种是栈顶符号是+或-，进去的为*或/，这种情况可以入栈。其他情况都可以把栈顶符号出栈。

只要理解符号的判断，其他的也不是难点。

if ((b == '*' || b =='/') && (a == '+' || a == '-')){
return true;
}
return false;

代码：

#include<iostream>
#include<cstring>
#include<stack>
using namespace std;
char s[205];
stack<double> stNum;
stack<char> stOp;

//   (1 + 2 ) * 3
//判断有括号的情况
bool check1(char b){
char a = stOp.top();
if (a == '(')
return true;
if ((b == '*' || b =='/') && (a == '+' || a == '-')){
return true;
}
return false;
}
//判断没有括号的情况
bool check1(char b){
char a = stOp.top();
if ((b == '*' || b == '/') && (a == '+' || a == '-')){
return true;
}
return false;
}
//用于没有括号
void solve(){
int len = strlen(s);
double sum;
for (int i = 0; i < len; i++){
if (s[i] == ' ') continue;
else if (s[i]=='+'||s[i]=='-'||s[i]=='*'||s[i]=='/'){
if (stOp.empty()){
stOp.push(s[i]);
}
else
{
while (stOp.size()>0&&!check2(s[i])){
double a = stNum.top(); stNum.pop();
double b = stNum.top(); stNum.pop();
char op = stOp.top(); stOp.pop();
double temp;
switch (op)
{
case '+':temp = a + b; break;
case '-':temp = b - a; break;
case '*':temp = a*b; break;
case '/':temp = b/a; break;
}
stNum.push(temp);
}
stOp.push(s[i]);
}
}
else if (s[i] >= '0'&&s[i] <= '9'){
sum = 0;
while (s[i] >= '0'&&s[i] <= '9')
{
sum *= 10;
sum += (s[i] - '0');
i++;
}
stNum.push(sum);
}
}
while (stOp.size()>0){
double a = stNum.top(); stNum.pop();
double b = stNum.top(); stNum.pop();
char op = stOp.top(); stOp.pop();
double temp;
switch (op)
{
case '+':temp = a + b; break;
case '-':temp = b - a; break;
case '*':temp = a*b; break;
case '/':temp = b / a; break;
}
stNum.push(temp);
}
printf("%.2lf\n", stNum.top());
stNum.pop();
}
//用于有括号
void solve2(){
int len = strlen(s);
double sum;
fo
4000
r (int i = 0; i < len; i++){
if (s[i] == ' ') continue;
else if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/'){
if (stOp.empty()){
stOp.push(s[i]);
}
else
{
while (stOp.size()>0 && !check1(s[i])){
double a = stNum.top(); stNum.pop();
double b = stNum.top(); stNum.pop();
char op = stOp.top(); stOp.pop();
double temp;
switch (op)
{
case '+':temp = a + b; break;
case '-':temp = b - a; break;
case '*':temp = a*b; break;
case '/':temp = b / a; break;
}
stNum.push(temp);
}
stOp.push(s[i]);
}
}
else if (s[i] >= '0'&&s[i] <= '9'){
sum = 0;
while (s[i] >= '0'&&s[i] <= '9')
{
sum *= 10;
sum += (s[i] - '0');
i++;
}
stNum.push(sum);
}
else if(s[i]=='('){
stOp.push(s[i]);
}
else if (s[i]==')')
{
while (stOp.top()!= '(')
{
double a = stNum.top(); stNum.pop();
double b = stNum.top(); stNum.pop();
char op = stOp.top(); stOp.pop();
double temp;
switch (op)
{
case '+':temp = a + b; break;
case '-':temp = b - a; break;
case '*':temp = a*b; break;
case '/':temp = b / a; break;
}
stNum.push(temp);
}
stOp.pop();
}
}
while (stOp.size()>0){
double a = stNum.top(); stNum.pop();
double b = stNum.top(); stNum.pop();
char op = stOp.top(); stOp.pop();
double temp;
switch (op)
{
case '+':temp = a + b; break;
case '-':temp = b - a; break;
case '*':temp = a*b; break;
case '/':temp = b / a; break;
}
stNum.push(temp);
}
printf("%.2lf\n", stNum.top());
stNum.pop();
}

int main(){
while (gets(s)!=NULL &&strcmp(s,"0")){
if (strlen(s) == 0) break;
solve();
}
return 0;
}