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题目链接：https://leetcode.com/problems/evaluate-reverse-polish-notation/
class Solution {
public:
int evalRPN(vector<string>& vec) {
stack<int> stk;
int len=vec.size();
int t=0,temp=0;
for(int i=0;i<len;i++)
{
if(vec[i]!="+"&&vec[i]!="-"&&vec[i]!="*"&&vec[i]!="/")
{
stk.push(stoi(vec[i]));
}
else if(vec[i]=="+")
{
t=stk.top();
stk.pop();
t+=stk.top();
stk.pop();
stk.push(t);
}
else if(vec[i]=="-")
{
t=stk.top();
stk.pop();
t=stk.top()-t;
stk.pop();
stk.push(t);
}
else if(vec[i]=="*")
{
t=stk.top();
stk.pop();
t=t*stk.top();
stk.pop();
stk.push(t);
}
else if(vec[i]=="/")
{
t=stk.top();
stk.pop();
t=stk.top()/t;
stk.pop();
stk.push(t);
}
}
return stk.top();
}
};