题目描述
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
//判断字符串是否是数字，如果是就压栈，不是就弹出2个元素进行计算，将计算结果压栈
//最后栈顶元素即为所求class Solution {
public:
int evalRPN(vector<string> &tokens) {
stack<int> st;
for (int i = ; i < tokens.size(); i++){
if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {
int temp = ;
int a = st.top();
st.pop();
int b = st.top();
st.pop();
if (tokens[i] == "+") { temp = b + a; }
else if (tokens[i] == "-") { temp = b - a; }
else if (tokens[i] == "*") { temp = b * a; }
else { temp = b / a; }
st.push(temp);
}
else {
st.push(stoi(tokens[i]));
}
}
return st.top();
}
};