Description
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
Solutions
1. Stack
# Time: O(n)
# Space: O(n)
class Solution:
def evalRPN(self, tokens: List[str]) -> int:
stack = []
operand = set({'+', '-', '*', '/'})
res = None
if len(tokens) == 1:
return int(tokens[0])
for ch in tokens:
if stack and ch in operand:
second = int(stack.pop())
if not stack:
return None
first = int(stack.pop())
if ch == '+':
tmp = first + second
elif ch == '-':
tmp = first - second
elif ch == '*':
tmp = first * second
else:
if (first >= 0 and second < 0) or (first <= 0 and second > 0):
tmp = abs(first) // abs(second)
tmp = -tmp
else:
tmp = first // second
res = tmp
stack.append(tmp)
else:
stack.append(ch)
return res
# 20/20 cases passed (68 ms)
# Your runtime beats 64.81 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.9 MB)