我正在编写一种解决各种方程的方法。现在我希望该方法接收一个可以采用以下形式的字符串方程：
ax^2+bx+c=0
或者
*ax^2+c=0*
或者
bx+c=0
等等，顺序应该无关紧要。
我的问题是：如何根据“x”等级解析方程？
eq 可以包含更多相同等级的值，例如 2x^2+4x^2+3x+8=2（最大等级 x^3）。
我的方法应该将 a 值分配给double a[]if 在 a 的左侧或右侧有x^2, double b[], 如果在左侧或右侧有x, 并且double c[]如果值附近没有任何x变量（并且应该更改值如果 therms 在=之后，请签名。
将字符串数字转换为双精度数很简单，但我不知道如何根据x描述的等级反汇编输入字符串。
经过测试-2x + 3x^2 - 2 + 3x = 3 - 2x^2
public Double[] parseEquation(String equation)
{
Log.d(TAG, "equation: " + equation);
// Remove all white spaces
equation = equation.replaceAll("[ ]", "");
// Get the left and right sides of =
String[] sides = equation.split("[=]"); // should be of size 2
boolean leftNegative = false;
boolean rightNegative = false;
if (sides.length != 2)
{
// There is no = or more than one = signs.
}
else
{
// if sides i starts with + remove the +
// if - we remove and put it back later
for (int i = 0; i < 2; i++)
{
if (sides[i].charAt(0) == '+')
{
sides[i] = sides[i].substring(1);
}
}
if (sides[0].charAt(0) == '-')
{
leftNegative = true;
sides[0] = sides[0].substring(1);
}
if (sides[1].charAt(0) == '-')
{
rightNegative = true;
sides[1] = sides[1].substring(1);
}
}
Log.d(TAG, "left side:" + sides[0] + " right side: " + sides[1]);
// Terms without signs need to find out later
String[] leftTerms = sides[0].split("[+-]");
String[] rightTerms = sides[1].split("[+-]");
int length = leftTerms[0].length();
if (leftNegative)
{
leftTerms[0] = "-" + leftTerms[0];
}
// put in the minus sign for the rest of the terms
for (int i = 1; i < leftTerms.length; i++)
{
Log.d(TAG, "length = " + length + " " + sides[0].charAt(length));
if (sides[0].charAt(length) == '-')
{
leftTerms[i] = "-" + leftTerms[i];
length += leftTerms[i].length();
}
else
{
length += leftTerms[i].length() + 1;
}
}
length = rightTerms[0].length();
if (rightNegative)
{
rightTerms[0] = "-" + rightTerms[0];
}
for (int i = 1; i < rightTerms.length; i++)
{
Log.d(TAG, "length = " + length + " " + sides[1].charAt(length));
if (sides[1].charAt(length) == '-')
{
rightTerms[i] = "-" + rightTerms[i];
length += rightTerms[i].length();
}
else
{
length += rightTerms[i].length() + 1;
}
}
// Now we put all the factors and powers in a list
List<ContentValues> leftLists = new ArrayList<ContentValues>();
// left side
for (int i = 0; i < leftTerms.length; i++)
{
Log.d(TAG, "leftTerm: " + leftTerms[i]);
ContentValues contentValues = new ContentValues();
int indexOfX = leftTerms[i].indexOf('x');
if (indexOfX == -1)
{
// no x mean a constant term
contentValues.put("factor", leftTerms[i]);
contentValues.put("power", "0");
}
else
{
int indexOfHat = leftTerms[i].indexOf('^');
if (indexOfHat == -1)
{
// no hat mean power = 1
contentValues.put("power", "1");
String factor = leftTerms[i].substring(0, indexOfX).trim();
contentValues.put("factor", factor);
}
else
{
String power = leftTerms[i].substring(indexOfX + 2).trim();
String factor = leftTerms[i].substring(0, indexOfX).trim();
contentValues.put("factor", factor);
contentValues.put("power", power);
}
}
Log.d(TAG, contentValues.toString());
leftLists.add(contentValues);
}
List<ContentValues> rightLists = new ArrayList<ContentValues>();
for (int i = 0; i < rightTerms.length; i++)
{
Log.d(TAG, "rightTerm: " + rightTerms[i]);
ContentValues contentValues = new ContentValues();
int indexOfX = rightTerms[i].indexOf('x');
if (indexOfX == -1)
{
// no hat mean a constant term
contentValues.put("factor", rightTerms[i]);
contentValues.put("power", "0");
}
else
{
int indexOfHat = rightTerms[i].indexOf('^');
if (indexOfHat == -1)
{
// no hat mean power = 1
contentValues.put("power", "1");
String factor = rightTerms[i].substring(0, indexOfX).trim();
contentValues.put("factor", factor);
}
else
{
String power = rightTerms[i].substring(indexOfX + 2).trim();
String factor = rightTerms[i].substring(0, indexOfX).trim();
contentValues.put("factor", factor);
contentValues.put("power", power);
}
}
Log.d(TAG, contentValues.toString());
rightLists.add(contentValues);
}
// Now add the factors with same powers.
// Suppose we solve for cubic here the end result will be
// 4 terms constant, x, x^2 and x^3
// Declare a double array of dim 4 the first will hold constant
// the second the x factor etc...
// You can allow arbitrary power by looping through the lists and get the max power
Double[] result = new Double[]{0.0, 0.0, 0.0, 0.0};
for (ContentValues c : leftLists)
{
switch (c.getAsInteger("power"))
{
case 0:
//Log.d(TAG, "power = 0, factor = " + c.toString());
result[0] += c.getAsDouble("factor");
break;
case 1:
result[1] += c.getAsDouble("factor");
break;
case 2:
result[2] += c.getAsDouble("factor");
break;
case 3:
result[3] += c.getAsDouble("factor");
break;
}
}
for (ContentValues c : rightLists)
{
switch (c.getAsInteger("power"))
{
case 0:
//Log.d(TAG, "power = 0, factor = " + c.toString());
result[0] -= c.getAsDouble("factor");
break;
case 1:
result[1] -= c.getAsDouble("factor");
break;
case 2:
result[2] -= c.getAsDouble("factor");
break;
case 3:
result[3] -= c.getAsDouble("factor");
break;
}
}
Log.d(TAG, "constant term = " + result[0] + ", x^1 = " + result[1]
+ ", x^2 = " + result[2] + ", x^3 = " + result[3]);
return result;
}
如果您不受 Android 的限制，我建议您使用词法分析器和解析器。这些是代码生成器，因此它们可以在基础语言工作的任何地方工作，但它们往往会产生臃肿的代码。Android 可能不会欣赏这一点。