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Given a string of an arithmetic expression; which includes only two operators: '+', '*' , write a function that evaluates the expression and returns its result.Input: A (possible) very long string of an arithmetic expression: "1*2*3+1+20".Output: 27
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| public class Solution { | |
| public static void main(String[] args) { | |
| String[] input = { "5", "4+20*3", "20*3+4", "20+20+20+4", "4*4*4" }; | |
| for (String test : input) | |
| System.out.printf("%s = %d%n", test, eval(test)); | |
| } | |
| private static int eval(String input) { | |
| int sum = 0; | |
| int multiplication = 1; | |
| String currentNumber = ""; | |
| if (input.matches("\\d+")) | |
| return Integer.parseInt(input); | |
| for (int i = 0; i < input.length(); i++) { | |
| char currentChar = input.charAt(i); | |
| switch (currentChar) { | |
| case '+': | |
| sum += multiplication * Integer.parseInt(currentNumber); | |
| multiplication = 1; | |
| currentNumber = ""; | |
| break; | |
| case '*': | |
| multiplication *= Integer.parseInt(currentNumber); | |
| currentNumber = ""; | |
| break; | |
| default: | |
| currentNumber += currentChar; | |
| break; | |
| } | |
| } | |
| return sum += multiplication * Integer.parseInt(currentNumber); | |
| } | |
| } |
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