//Recursive
int evalRPN(vector<string> &n) {
string s = n.back(); n.pop_back();
if ( s== "*" || s=="/" || s=="+" || s == "-" ){
int r2 = evalRPN(n);
int r1 = evalRPN(n);
if ( s=="*") return r1*r2;
if ( s=="/") return r1/r2;
if ( s=="+") return r1+r2;
if ( s=="-") return r1-r2;
}
else
return atoi(s.c_str());
}
int evalRPN(vector<string> &tokens) {
int x, y;
auto token = tokens.back(); tokens.pop_back();
if (is_operator(token)) {
y = evalRPN(tokens);
x = evalRPN(tokens);
if (token[0] == '+') x += y;
else if (token[0] == '-') x -= y;
else if (token[0] == '*') x *= y;
else x /= y;
} else {
size_t i;
x = stoi(token, &i);
}
return x;
}
double evaluate(const std::string &expr) {
std::queue<std::string> output;
parse(expr, output);
double result = evalRPN(output);
return result;
}
int evalRPN(vector<string> &tokens) {
if(!tokens.size()) return 0;
if(tokens.size() == 1)
return stoi(tokens.front());
string op;
int num1, num2;
op = tokens.back();
tokens.pop_back();
int idx = tokens.size(), expectNum = 1;
vector<string>::iterator it = tokens.end();
vector<string> aux;
while(true){
it--;
idx--;
if(isOperator(tokens[idx])){
expectNum += 1;
}else if(--expectNum == 0){
aux.insert(aux.begin(), it, tokens.end());
tokens.erase(it, tokens.end());
num1 = evalRPN(tokens);
num2 = evalRPN(aux);
break;
}
}
if(op == "+"){
return num1 + num2;
}else if(op == "-"){
return num1 - num2;
}else if(op == "*"){
return num1 * num2;
}else if(op == "/"){
return num1 / num2;
}else{
cout << "wrong operator\n";
}
}
int evalRPN(vector<string> &tokens) {
stack<int> s;
for (vector<string>::const_iterator it = tokens.begin(); it != tokens.end(); ++it) {
//如果是数字
if (isNumber(*it))
s.push(stoi(*it)); //将数字压入栈中
//如果是运算符
else if (*it == "+" || *it == "-" || *it == "*" || *it == "/") {
if (!s.empty()) { //如果栈中有数字
int opnd2 = s.top(); //获得表达式中的右边的操作数
s.pop(); //弹出
int opnd1 = s.top(); //获得左边的操作数
s.pop(); //弹出
s.push(calculate(opnd1, it->c_str(), opnd2)); //计算表达式，将结果再压入栈中
}
}
//如果是表达式
else {
vector<string> anotherTokens;
s.push(evalRPN(anotherTokens = convertToRPN(*it))); //如果是表达式那么转化成逆波兰式，在调用evalRPN，所得结果压入栈中
}
}
return s.top(); //返回最终结果
}
void test() {
vector<string> tokens = { "-1", "1", "*", "-1", "+" };
int result = evalRPN(tokens);
}
int calculate(string s)
{
vector <string> sets = postfix(s);
return evalRPN(sets);
}