[英]Java Program Optimization Involving String Comparison
因此,我制作了一個簡單的程序,該程序測試兩個字符串值(代表兩種單獨的原色)的區別和類型,以確定所得混合物的顏色。
/**
* A program that prompts the user to enter the names of two different primary colors to create a mixture.
*
* @author A. Mackey
* @version 1.0
*/
import java.util.*;
public class ColourMixer {
public static void main(String [] args) {
colourMixer();
}
public static void colourMixer() {
Scanner in = new Scanner(System.in);
String colourOne = "";
String colourTwo = "";
System.out.println("You are mixing two different primary colours.");
System.out.print("Enter your first colour: ");
colourOne = in.nextLine();
System.out.print("Enter your second colour: ");
colourTwo = in.nextLine();
if((colourOne.equalsIgnoreCase("red") || colourOne.equalsIgnoreCase("blue") ) && (colourTwo.equalsIgnoreCase("blue") || colourTwo.equalsIgnoreCase("red") && !(colourOne.equalsIgnoreCase(colourTwo)))) {
System.out.println("Your colour combination creates purple!");
} else if ((colourOne.equalsIgnoreCase("red") || colourOne.equalsIgnoreCase("yellow") ) && (colourTwo.equalsIgnoreCase("yellow") || colourTwo.equalsIgnoreCase("red") && !(colourOne.equalsIgnoreCase(colourTwo)))) {
System.out.println("Your colour combination creates orange!");
} else if ((colourOne.equalsIgnoreCase("blue") || colourOne.equalsIgnoreCase("yellow") ) && (colourTwo.equalsIgnoreCase("yellow") || colourTwo.equalsIgnoreCase("blue") && !(colourOne.equalsIgnoreCase(colourTwo)))) {
System.out.println("Your colour combination creates green!");
} else
System.out.println("You have not entered two different primary colours.");
}
}
它的功能完全符合預期,沒有明顯的問題。
但是,它在邏輯上感到笨拙和過度。
我很好奇這個算法是否可以改進。
也許一些布爾值刪除重復項。
boolean isRed = colourOne.equalsIgnoreCase("red") || colourTwo.equalsIgnoreCase("red") ;
boolean isBlue = colourOne.equalsIgnoreCase("blue") || colourTwo.equalsIgnoreCase("blue") ;
if (isRed && isBlue){
// ...
} else if (isRed && isYellow) {
// ...
您可以將輸入字符串(在我的情況下為AS小寫字符串)放入Set中,並可以調用Sets contains。
如果setContains(red)->檢查是否包含黃色或藍色。 現在您可以選擇setContains(blue)或setContains(yellow)並測試是否存在其他一種顏色,因為通過這種方式,顏色的組合無關緊要,並且測試了所有可能的情況。 黃色和藍色將與藍色和黃色相同。
Set<String> set = new HashSet<String>();
set.add(colourOne.toLowerCase());
set.add(colourTwo.toLowerCase());
if(set.contains("red")){
if(set.contains("blue")){
//return purple
}
if(set.contains("yellow"){
//returnorange
}
}
if(set.contains("yellow")){
if(set.contains("blue"){
//return green
}
}
//unidentified colour
可用的另一個選項是將它們存儲在列表中。 然后,您可以檢查列表中是否有顏色。
例如
String colourOne = "blue";
String colourTwo = "red";
List<String> colours= new ArrayList<String>(2);
colours.add(colourOne);
colours.add(colourTwo);
if((colours.contains("red") && colours.contains("blue")
System.out.println("Your colour combination creates purple!");
etc
這看起來像是使用枚舉的理想場所。
enum COLOR{
RED,GREEN,BLUE,BLACK,WHITE; //etc.
}
然后,您可以使用一種方法將用戶輸入的字符串轉換為顏色:
public static final COLOR convertToColor(final String input){
try{
return COLOR.valueOf(input.toUpperCase());
}catch(IllegalArgumentException e){
//invalid color, handle it properly
System.out.println(input + " is not a valid color !! Enter a new valid color this time!!");
return null;
}
}
public static void colourMixer() {
Scanner in = new Scanner(System.in);
COLOR color1;
COLOR color2;
System.out.print("Enter your first colour: ");
do{
color1 = convertToColor(in.nextLine());
}while( color1 == null);
System.out.print("Enter your second colour: ");
do{
color2 = convertToColor(in.nextLine());
}while( color2 == null);
if(COLOR.RED.equals(color1) && COLOR.BLUE.equals(color2)){
System.out.println("Your colour combination creates purple!");
}else if(COLOR.RED.equals(color1) && COLOR.YELLOW.equals(color2)){
System.out.println("Your colour combination creates orange!");
}//....
}
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