逆波兰表达式

利用栈，解决逆波兰表达式

//逆波兰表达式
bool IsOperator(const char* token)
{
    return ((token[0] == '+') || (token[0] == '-')|| (token[0]=='*')||(token[0]=='/'));
}

int ReversePolishNotation(const char* str[],int size)
{
    stack<int> s;
    int a,b;
    const char* token;
    for (int i = 0; i<size; i++) {
        token = str[i];
        if (!IsOperator(token)) {
            s.push(atoi(token));    //int atoi(const char *nptr);将字符转化为整型
        }
        else
        {
            b = s.top();
            s.pop();
            a = s.top();
            s.pop();
            if (token[0]=='+') {
                s.push(a+b);
            }
            else if(token[0]=='-'){
                s.push(a-b);
            }
            else if (token[0]=='*'){
                s.push(a*b);
            }
            else
                s.push(a/b);
        }
    }
    return s.top();
}

int main()
{
    const char* str[]={"2","1","+","3","*"};
    int value = ReversePolishNotation(str,sizeof(str)/sizeof(const char*));
    cout<<value<<endl;
    return 0;
}