Problem
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
Solution
/**
* @param {string[]} tokens
* @return {number}
*/
var evalRPN = function(tokens) {
var stack = [];
var len = tokens.length;
var n1 = 0;
var n2 = 0;
var n3 = 0;
var token = '';
for (var i = 0; i < len; i++) {
token = tokens[i];
switch (token) {
case '+':
stack.push(stack.pop() + stack.pop());
break;
case '-':
n1 = stack.pop();
n2 = stack.pop();
stack.push(n2 - n1);
break;
case '*':
stack.push(stack.pop() * stack.pop());
break;
case '/':
n1 = stack.pop();
n2 = stack.pop();
n3 = n2 / n1;
stack.push(n3 > 0 ? Math.floor(n3) : Math.ceil(n3));
break;
default:
stack.push(Number(token));
}
}
return stack.pop();
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).