150J. Evaluate Reverse Polish Notation
difficulty: Medium
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +
, -
, *
, and /
. Each operand may be an integer or another expression.
Note that division between two integers should truncate toward zero.
It is guaranteed that the given RPN expression is always valid. That means the expression would always evaluate to a result, and there will not be any division by zero operation.
Example 1:
Input: tokens = ["2","1","+","3","*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9
Example 2:
Input: tokens = ["4","13","5","/","+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6
Example 3:
Input: tokens = ["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
Constraints:
1 <= tokens.length <= 104
tokens[i]
is either an operator:"+"
,"-"
,"*"
, or"/"
, or an integer in the range[-200, 200]
.
Method One
class Solution {
public int evalRPN(String[] tokens) {
ArrayDeque<Integer> values = new ArrayDeque<>();
for(String str : tokens){
if(str.equals("+")){
values.push(values.pop() + values.pop());
}else if(str.equals("-")){
values.push(- values.pop() + values.pop());
}else if(str.equals("*")){
values.push(values.pop() * values.pop());
}else if(str.equals("/")){
int a = values.pop();
int b = values.pop();
values.push(b/a);
}else{
values.push(Integer.valueOf(str));
}
}
return values.pop();
}
}
202309写的,很多年过去了这也太巧了吧,和之前的几乎一样。
class Solution {
public int evalRPN(String[] tokens) {
ArrayDeque<Integer> stack = new ArrayDeque<>();
for (String token : tokens) {
switch (token) {
case "+":
stack.push(stack.pop() + stack.pop());
break;
case "-":
stack.push( - stack.pop() + stack.pop());
break;
case "*":
stack.push(stack.pop() * stack.pop());
break;
case "/":
int b = stack.pop();
int a = stack.pop();
stack.push(a / b);
break;
default:
stack.push(Integer.valueOf(token));
}
}
return stack.pop();
}
}
Last updated
Was this helpful?