Using the "stack" fast computing -- reverse polish expression

Example: a b c * + d e * f + g * +

Corresponding data: 1 2 3 * + 4 5 * 6 + 7 * +

Analysis: (the data processing area is a virtual module)

The last element left on the stack at this point is the result of the expression

Example: Reverse Polish Expression Evaluation

Based on the above analysis, it will be much easier to do this question~ The code is as follows:

class Solution {public int evalRPN(String[] tokens) {Stack stack = new Stack<>();for(String str : tokens){if(judgment(str)){int x = stack.pop();int y = stack.pop();switch(str){case "+":stack.push(y + x);break;case "-":stack.push(y - x);break;case "*":stack.push(y * x);break;case "/":stack.push(y/x);break;}}else{stack.push(Integer.parseInt(str));}}return stack.pop();}private boolean judgment(String str){if(str.equals("+") || str.equals("-") ||str.equals("*") || str.equals("/")){return true;}else{return false;}}}

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