[英]parsing a quadratic equation in java
我必須為一個二次類編寫一個read方法,其中以ax ^ 2 + bx + c = 0的形式輸入二次方,我發現這種方式,這是代碼:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class ParseEquation {
public static String coeff(String str, String regex) {
Pattern patt = Pattern.compile(regex);
Matcher match = patt.matcher(str);
// missing coefficient default
String coeff = "+0";
if(match.find())
coeff = match.group(1);
// always have sign, handle implicit 1
return (coeff.length() == 1) ? coeff + "1"
: coeff;
}
public static String[] quadParse(String arg) {
String str = ("+" + arg).replaceAll("\\s", "");
String a = coeff(str, "([+-][0-9]*)x\\^2" );
String b = coeff(str, "([+-][0-9]*)x(?!\\^)");
String c = coeff(str, "([+-][0-9]+)(?!x)" );
double a1 = Double.parseDouble(a);
double b1 = Double.parseDouble(b);
double c1 = Double.parseDouble(c);
double dis = (Math.pow(b1, 2.0)) - (4 * a1 * c1);
double d = Math.sqrt(dis);
double X = 0,Y = 0; //root 1 & root 2, respectively
if (dis > 0.0 || dis < 0.0 ) {
X = (-b1 + d)/(2.0 * a1 );
Y = (-b1 - d)/(2.0 *a1);
String root1 = Double.toString(X);
String root2 = Double.toString(Y);
return new String[]{root1,root2};
} else if (dis == 0.0){
X = (-b1 + 0.0)/(2.0 * a1);//repeated root
String root2 = Double.toString(X);
return new String[]{root2};
}
return new String[-1];
}
public static void main(String[] args) throws IOException {
BufferedReader r = new BufferedReader (new InputStreamReader (System.in));
String s;
while ((s=r.readLine()) != null) {
String[] pieces = quadParse(s);
System.out.println(Arrays.toString(pieces));
}
}
}
這已經解決了正常的二階方程ax ^ 2 + bx + c = 0(例如,用戶輸入的任何方程式,例如“ 2x ^ 2 + 2x -25 = 0”),並且可以通過改變根的位置來求解,但是我不知道如何求解具有類似根的方程,例如“ 2x ^ 2 + 2x -3x -25 +15 = 0”,因此應首先將x的兩個系數求和,然后求和(-25 + 15),然后計算結果。 所以我需要知道知道如何編寫代碼來做到這一點的方法。
隨意編寫任何代碼作為示例。
我認為您需要不斷將x^2
和x
的值相加。 我已經修改了代碼,而且看起來工作正常:
public class ParseEquation {
public static double coeff(String str, String regex) {
Pattern patt = Pattern.compile(regex);
Matcher match = patt.matcher(str);
// missing coefficient default
String coeff = "+0";
double value = 0;
while(match.find()){
coeff = match.group(1);
value = value + Double.parseDouble(coeff);
}
// always have sign, handle implicit 1
return (coeff.length() == 1) ? (value + 1) : value;
}
public static String[] quadParse(String arg) {
String str = ("+" + arg).replaceAll("\\s", "");
double a1 = coeff(str, "([+-][0-9]*)x\\^2");
double b1 = coeff(str, "([+-][0-9]*)x(?!\\^)");
double c1= coeff(str, "([+-][0-9]+)(?!x)");
System.out.println("Values are a: " + a1 + " b: " + b1 + " c: " + c1);
double dis = (Math.pow(b1, 2.0)) - (4 * a1 * c1);
double d = Math.sqrt(dis);
double X = 0, Y = 0; //root 1 & root 2, respectively
if (dis > 0.0 || dis < 0.0) {
X = (-b1 + d) / (2.0 * a1);
Y = (-b1 - d) / (2.0 * a1);
String root1 = Double.toString(X);
String root2 = Double.toString(Y);
return new String[]{root1, root2};
} else if (dis == 0.0) {
X = (-b1 + 0.0) / (2.0 * a1);//repeated root
String root2 = Double.toString(X);
return new String[]{root2};
}
return new String[-1];
}
public static void main(String[] args) throws IOException {
BufferedReader r = new BufferedReader (new InputStreamReader(System.in));
String s;
while ((s=r.readLine()) != null) {
String[] pieces = quadParse(s);
System.out.println(Arrays.toString(pieces));
}
}
}
這是我運行程序時的輸出:
2x^2 + 2x -3x -25 +15 =0 Values are a: 2.0 b: -1.0 c: -10.0 [2.5, -2.0]
因此,它能夠正確地求和系數。 我沒有改變您編寫的IMO應該能正常工作的邏輯。
編輯以使其處理隱式后的最終答案1
例如:x ^ 2-3x-2x-25
值是a:1.0 b:-5.0 c:-25.0
[8.090169943749475,-3.0901699437494745]
public class ParseEquation {
public static String coeff(String str, String regex) {
Pattern patt = Pattern.compile(regex);
Matcher match = patt.matcher(str);
// missing coefficient default
String coeff = "+0";
double value = 0;
if(match.find())
coeff = match.group(1);
// always have sign, handle implicit 1
value= Double.parseDouble((coeff.length() == 1) ? coeff + "1"
: coeff);
while(match.find()){
coeff = match.group(1);
value = value + Double.parseDouble(coeff);
}
String value2 =String.valueOf(value);
return (value2.length() == 1) ? (value2 + "1") : value2;
}
public static String[] quadParse(String arg) {
String str = ("+" + arg).replaceAll("\\s", "");
double a1 = Double.parseDouble(coeff(str, "([+-][0-9]*)([a-z]\\^2)"));
double b1 = Double.parseDouble(coeff(str, "([+-][0-9]*)([a-z](?!\\^))"));
double c1= Double.parseDouble(coeff(str, "([+-][0-9]+)(?![a-z])"));
System.out.println("Values are a: " + a1 + " b: " + b1 + " c: " + c1);
double dis = (Math.pow(b1, 2.0)) - (4 * a1 * c1);
double d = Math.sqrt(dis);
double X = 0, Y = 0; //root 1 & root 2, respectively
if (dis > 0.0 || dis < 0.0) {
X = (-b1 + d) / (2.0 * a1);
Y = (-b1 - d) / (2.0 * a1);
String root1 = Double.toString(X);
String root2 = Double.toString(Y);
return new String[]{root1, root2};
} else if (dis == 0.0) {
X = (-b1 + 0.0) / (2.0 * a1);//repeated root
String root2 = Double.toString(X);
return new String[]{root2};
}
return new String[-1];
}
public static void main(String[] args) throws IOException {
BufferedReader r = new BufferedReader (new InputStreamReader(System.in));
String s;
while ((s=r.readLine()) != null) {
String[] pieces = quadParse(s);
System.out.println(Arrays.toString(pieces));
}
}
}
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