Java中BODMAS的算法计算
我已经为Bodmas编写了这段代码,但是在这段代码中出现了一些错误。如果我做3-5+9,结果将是3.04.0 它只是开始连接,尽管它适用于所有其他操作,如*、/和-,请帮助Java中BODMAS的算法计算,java,Java,我已经为Bodmas编写了这段代码,但是在这段代码中出现了一些错误。如果我做3-5+9,结果将是3.04.0 它只是开始连接,尽管它适用于所有其他操作,如*、/和-,请帮助 public static String calculation(BODMASCalculation bodmas, String result) { while (bodmas.hasMatch()) { double value, leftOfOperator = bodmas.getLeft()
public static String calculation(BODMASCalculation bodmas, String result) {
while (bodmas.hasMatch()) {
double value, leftOfOperator = bodmas.getLeft();
char op = bodmas.getOperator();
double rightOfOprator = bodmas.getRight();
switch (op) {
case '/':
if(rightOfOprator == 0) //Divide by 0 generates Infinity
value = 0;
else
value = leftOfOperator / rightOfOprator;
break;
case '*':
value = leftOfOperator * rightOfOprator;
break;
case '+':
value = leftOfOperator + rightOfOprator;
break;
case '-':
value = leftOfOperator - rightOfOprator;
break;
default:
throw new IllegalArgumentException("Unknown operator.");
}
result = result.substring(0, bodmas.getStart()) + value + result.substring(bodmas.getEnd());
bodmas = new BODMASCalculation(result);
}
return result;
}
另一个功能是:-
public boolean getMatchFor(String text, char operator) {
String regex = "(-?[\\d\\.]+)(\\x)(-?[\\d\\.]+)";
java.util.regex.Matcher matcher = java.util.regex.Pattern.compile(regex.replace('x', operator)).matcher(text);
if (matcher.find()) {
this.leftOfOperator = Double.parseDouble(matcher.group(1));
this.op = matcher.group(2).charAt(0);
this.rightOfOprator = Double.parseDouble(matcher.group(3));
this.start = matcher.start();
this.end = matcher.end();
return true;
}
return false;
}
我有一个解决方案,添加
String sss = null;
if(op == '+' && !Str.isBlank(result.substring(0, bodmas.getStart())) && value >= 0)
sss = "+";
else
sss = "";
result = result.substring(0, bodmas.getStart()) + sss + value + result.substring(bodmas.getEnd());
但我不想这样做,我希望这对所有的操作符都有效。import java.util.Stack;
import java.util.Stack;
public class EvaluateString
{
public static int evaluate(String expression)
{
char[] tokens = expression.toCharArray();
// Stack for numbers: 'values'
Stack<Integer> values = new Stack<Integer>();
// Stack for Operators: 'ops'
Stack<Character> ops = new Stack<Character>();
for (int i = 0; i < tokens.length; i++)
{
// Current token is a whitespace, skip it
if (tokens[i] == ' ')
continue;
// Current token is a number, push it to stack for numbers
if (tokens[i] >= '0' && tokens[i] <= '9')
{
StringBuffer sbuf = new StringBuffer();
// There may be more than one digits in number
while (i < tokens.length && tokens[i] >= '0' && tokens[i] <= '9')
sbuf.append(tokens[i++]);
values.push(Integer.parseInt(sbuf.toString()));
}
// Current token is an opening brace, push it to 'ops'
else if (tokens[i] == '(')
ops.push(tokens[i]);
// Closing brace encountered, solve entire brace
else if (tokens[i] == ')')
{
while (ops.peek() != '(')
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
ops.pop();
}
// Current token is an operator.
else if (tokens[i] == '+' || tokens[i] == '-' ||
tokens[i] == '*' || tokens[i] == '/')
{
// While top of 'ops' has same or greater precedence to current
// token, which is an operator. Apply operator on top of 'ops'
// to top two elements in values stack
while (!ops.empty() && hasPrecedence(tokens[i], ops.peek()))
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
// Push current token to 'ops'.
ops.push(tokens[i]);
}
}
// Entire expression has been parsed at this point, apply remaining
// ops to remaining values
while (!ops.empty())
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
// Top of 'values' contains result, return it
return values.pop();
}
// Returns true if 'op2' has higher or same precedence as 'op1',
// otherwise returns false.
public static boolean hasPrecedence(char op1, char op2)
{
if (op2 == '(' || op2 == ')')
return false;
if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-'))
return false;
else
return true;
}
// A utility method to apply an operator 'op' on operands 'a'
// and 'b'. Return the result.
public static int applyOp(char op, int b, int a)
{
switch (op)
{
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
if (b == 0)
throw new
UnsupportedOperationException("Cannot divide by zero");
return a / b;
}
return 0;
}
// Driver method to test above methods
public static void main(String[] args)
{
System.out.println(EvaluateString.evaluate("10 + 2 * 6"));
System.out.println(EvaluateString.evaluate("100 * 2 + 12"));
System.out.println(EvaluateString.evaluate("100 * ( 2 + 12 )"));
System.out.println(EvaluateString.evaluate("100 * ( 2 + 12 ) / 14"));
}
}
公共类测试环
{
公共静态整型求值(字符串表达式)
{
char[]tokens=expression.toCharArray();
//数字堆栈:“值”
堆栈值=新堆栈();
//操作员堆栈:“ops”
堆栈操作=新堆栈();
for(int i=0;i 如果(令牌[i]>='0'&&tokens[i]='0'&&tokens[i]Java脚本API允许您将参数从Java应用程序传递到脚本引擎,反之亦然
您可以使用Javax ScriptEngine将值从应用程序传递到脚本
使用它的eval()
方法,你可以给它一个字符串形式的数学表达式,它将为你做数学运算…(也处理BODMAS)
例如:
输出:70
如果IDE不建议,请包括以下导入:
import javax.script.ScriptEngineManager;
import javax.script.ScriptEngine;
import javax.script.ScriptException;
请检查此处的文档
我希望它能有所帮助。“BODMAS”不是一个很实用的规则。尤其是加法和减法具有相同的优先级,从左到右计算1-2+3-4+5=((1-2)+3)-4)+5
该规则适用于嵌套循环
环路
替换所有(
编号)
-->编号
替换所有编号“[*/]”编号-->编号操作编号
替换所有编号“[+-]”编号-->编号操作编号
直到什么都没有被取代
这确保了3-4/26+5-2->3-26+5-2->3-12+5-3->-9+5-3->-4您尝试过使用调试器吗?如果是,您从使用调试器中学到了什么?是的,我可以修复,但它不会是通用的。所以我需要一个通用的调试器。修复什么?您所说的“通用”是什么意思?适用于所有算术运算的代码。不要将代码放在注释中。它几乎不可读。还请为您的答案添加一些解释。仅代码不能解释任何内容,被认为是低质量的。请查看和您的答案。这是Dijkstra调车场算法。您当然应该这么说,您应该已经解释了这是如何解决问题的,以及为什么会解决这个问题。此外,这不适用于负输入的情况,例如(“-4*-4”)
import javax.script.ScriptEngineManager;
import javax.script.ScriptEngine;
import javax.script.ScriptException;