簡單的四則運算電腦

   這是一個簡單的四則運算電腦，不支援括号，沒有做乘法的越界計算和除0等異常處理。

   隻是簡單的描述一個四則運算過程。

package SortSet;

import java.util.Stack;

public class OperationTest {

	private static Stack<Character> opeStack = new Stack<Character>();
	private static Stack<Integer> numStack = new Stack<Integer>();

	private static StringBuilder in = new StringBuilder();
	private static StringBuilder number = new StringBuilder();

	public static void main(String[] args) {
		System.out.println(cal("4-6*8-5+2*11"));
	}

	public static int cal(String input) {
		in.append(input);
		while (in.length() > 0) {
			char temp = in.charAt(0);
			in.delete(0, 1);
			if (temp > '0' && temp < '9') {
				number.append(temp);	
			} else {
				if (!"".equals(number.toString())) {
					int num = Integer.parseInt(number.toString());
					numStack.push(num);
					number.delete(0, number.length());
				}
				while (!opeStack.isEmpty() && !compare(temp)) {
					int num2 = numStack.pop();
					int num1 = numStack.pop();
					char ope = opeStack.pop();
					cal0(ope,num1,num2);
				}
				opeStack.push(temp);	
			}
		}
		if(number.length()>0){
			int num = Integer.parseInt(number.toString());
			numStack.push(num);
		}
		while(!opeStack.isEmpty()){
			int num2= numStack.pop();
			int num1= numStack.pop();
			cal0(opeStack.pop(),num1,num2); 
		}
			
		return numStack.pop();
	}

	public static void cal0(char ope,int num1 ,int num2){
		switch (ope) {
		case '+':
			numStack.push(num1 + num2);
			break;
		case '-':
			numStack.push(num1 - num2);
			break;
		case '*':
			numStack.push(num1 * num2);
			break;
		case '/':
			numStack.push(num1 / num2);
			break;
		}
	}
	public static boolean compare(char operation) {
		if (opeStack.isEmpty())
			return true;
		char last = opeStack.peek();

		switch (operation) {
		case '+':
			return false;
		case '-':
			return false;
		case '*':
			if (last == '+' || last == '-')
				return true;
			return false;
		case '/':
			if (last == '+' || last == '-')
				return true;
			return false;
		}
		return true;
	}

}
           

 執行結果為：

-27

說明：這裡乘法問題需要考慮越界，最好用大整數得結果。另外除法也需要考慮除0異常。 這裡采用StringBuilder減少了建立String的開銷。 算法過程為将操作數壓入操作棧，如果目前操作符優先級低于棧頂優先級，則計算，否則壓入操作符棧。當最終計算式周遊完時，進行運算即可。