# 150. Evaluate Reverse Polish Notation
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Note:
- Division between two integers should truncate toward zero. *The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.
Example 1:
Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9
Example 2:
Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6
Example 3:
Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation:
((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22
# Solution
Approach 1: Stack.
# Code (Python)
Approach 1:
# Code (C++)
Approach 1:
class Solution {
private:
int getNum(string s) {
int head = 0;
int sign = 1;
if (s[0] == '+' || s[0] == '-')
{
if (s[0] == '-')
sign = -1;
head++;
}
int num = 0;
for (int i = head; i < s.size(); ++i)
{
num = num * 10 + (s[i] - '0');
}
return sign * num;
}
int doCalculation(int a, int b, char op) {
switch (op)
{
case '+' : return a + b;
case '-' : return a - b;
case '*' : return a * b;
case '/' : return a / b;
}
return 0;
}
public:
int evalRPN(vector<string>& tokens) {
stack<int> st;
for (int i = 0; i < tokens.size(); ++i)
{
string s = tokens[i];
if (s == "+" || s == "-" || s == "*" || s == "/")
{
int num2 = st.top();
st.pop();
int num1 = st.top();
st.pop();
st.push(doCalculation(num1, num2, s[0]));
}
else
{
// st.push(getNum(s));
// or
st.push(atoi(s.c_str()));
}
}
return st.top();
}
};