Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. For example:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
1. Naive Approach
This problem can be solved by using a stack. We can loop through each element in the given array. When it is a number, push it to the stack. When it is an operator, pop two numbers from the stack, do the calculation, and push back the result.
public class Test {public static void main(String[] args) throws IOException {String[] tokens = new String[] { "2", "1", "+", "3", "*" };System.out.println(evalRPN(tokens));}public static int evalRPN(String[] tokens) {int returnValue = 0;String operators = "+-*/";Stack<String> stack = new Stack<String>();for (String t : tokens) {if (!operators.contains(t)) { //push to stack if it is a numberstack.push(t);} else {//pop numbers from stack if it is an operatorint a = Integer.valueOf(stack.pop());int b = Integer.valueOf(stack.pop());switch (t) {case "+":stack.push(String.valueOf(a + b));break;case "-":stack.push(String.valueOf(b - a));break;case "*":stack.push(String.valueOf(a * b));break;case "/":stack.push(String.valueOf(b / a));break;}}}returnValue = Integer.valueOf(stack.pop());return returnValue;}}
or
public class Solution {public int evalRPN(String[] tokens) {int returnValue = 0;String operators = "+-*/";Stack<String> stack = new Stack<String>();for(String t : tokens){if(!operators.contains(t)){stack.push(t);}else{int a = Integer.valueOf(stack.pop());int b = Integer.valueOf(stack.pop());int index = operators.indexOf(t);switch(index){case 0:stack.push(String.valueOf(a+b));break;case 1:stack.push(String.valueOf(b-a));break;case 2:stack.push(String.valueOf(a*b));break;case 3:stack.push(String.valueOf(b/a));break;}}}returnValue = Integer.valueOf(stack.pop());return returnValue;}}
时间: 2024-10-22 21:00:59