Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +-*/. Each operand may be an integer or another expression.

Some examples:

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6 此题是求解后缀表达式,题目比较简单,开一个数据stack即可,此题的改进版本是引入括号,此时要开一个数据stack和运算符stack
bool isOperator(string& op){
if(op == "+" || op == "-" || op == "*" || op == "/") return true;
else return false;
} int calc(int a, int b, char op){
switch(op){
case '+':return a+b;
case '-':return a-b;
case '*':return a*b;
case '/':return a/b;
}
} int evalRPN(vector<string> &tokens){
stack<int> num;
for(int i = ; i < tokens.size(); ++ i){
if(!isOperator(tokens[i])){
num.push(atoi(tokens[i].c_str()));
}else{
int num2 = num.top(); num.pop();
int num1 = num.top(); num.pop();
char op = tokens[i][];
num.push(calc(num1,num2,op));
}
}
return num.top();
}
 

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