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:

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Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Example 2:

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Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Example 3:

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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

  • 使用栈能够简单的实现
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class  {
public int evalRPN(String[] tokens) {
Stack<Integer> s = new Stack<>();
for (String t: tokens){
if ("+".equals(t)){
int b = s.pop();
int a = s.pop();
int ans = a + b;
s.push(ans);
}
else if ("-".equals(t)){
int b = s.pop();
int a = s.pop();
int ans = a - b;
s.push(ans);

}
else if ("*".equals(t)){
int b = s.pop();
int a = s.pop();
int ans = a * b;
s.push(ans);
}
else if ("/".equals(t)){
int b = s.pop();
int a = s.pop();
int ans = a / b;
s.push(ans);
}
else{
int num = Integer.parseInt(t);
s.push(num);
}
}
return s.peek();
}
}
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