150. Evaluate Reverse Polish Notation (Medium)
https://leetcode.com/problems/evaluate-reverse-polish-notation/
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
class Solution {
// Dead simple
public int evalRPN(String[] tokens) {
if (tokens == null || tokens.length == 0) {
return 0;
}
Stack<Integer> stack = new Stack<>();
for (int i = 0; i < tokens.length; i++) {
if (tokens[i].equals("+")) {
int val1 = stack.pop();
int val2 = stack.pop();
stack.push(val2 + val1);
continue;
}
if (tokens[i].equals("-")) {
int val1 = stack.pop();
int val2 = stack.pop();
stack.push(val2 - val1);
continue;
}
if (tokens[i].equals("*")) {
int val1 = stack.pop();
int val2 = stack.pop();
stack.push(val2 * val1);
continue;
}
if (tokens[i].equals("/")) {
int val1 = stack.pop();
int val2 = stack.pop();
stack.push(val2 / val1);
continue;
}
stack.push(Integer.valueOf(tokens[i]));
}
return stack.pop();
}
}