简单枚举类型——植物与颜色
Time Limit: 1000 ms
Memory Limit: 65536 KiB
Problem Description
请定义具有red, orange, yellow, green, blue, violet六种颜色的枚举类型color，根据输入的颜色名称，输出以下六种植物花朵的颜色：
Rose(red), Poppies(orange), Sunflower(yellow), Grass(green), Bluebells(blue), Violets(violet)。如果输入的颜色名称不在枚举类型color中，例如输入purple，请输出I don't know about the color purple.
Input
输入数据有多行，每行有一个字符串代表颜色名称，颜色名称最多30个字符，直到文件结束为止。
Output
输出对应颜色的植物名称，例如：Bluebells are blue. 如果输入的颜色名称不在枚举类型color中，例如purple, 请输出I don't know about the color purple.
Sample Input
blue yellow purple
Sample Output
Bluebells are blue. Sunflower are yellow. I don't know about the color purple.
package hello;
import java.util.Scanner;
enum Color{
red("Rose", "red"), //建立对应的枚举类型
orange("Poppies", "orange"),
yellow("Sunflower","yellow"),
green("Grass", "green"),
blue("Bluebells", "blue"),
violet("Violets", "violet");
String name;
String color;
Color(String name, String color){ //构造方法
this.name = name;
this.color = color;
}
public String toString() { //重写toSting方法
return name+" are "+color+".";
}
}
public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int flag = 1;
while(cin.hasNext()) {
String incolor = cin.next();
/*Color color = Color.valueOf(incolor); //本来想通过转化之后判断一个该颜色有没有在枚举序列中，当时如果没在序列中
for(Color a:color.values()) { //valueOf函数就会抛出异常，程序没法执行，所以用try来实现
if(a.equals(color)) {
flag = 0;
break;
}
}
if(flag == 0) {
System.out.println(color);
}
else {
System.out.println("I don't know about the color "+incolor+".");
}*/
try{ //如果输入的颜色不在对应的颜色序列当中，就会抛出异常
Color color = Color.valueOf(incolor);
System.out.println(color);
}
catch(Exception e){
System.out.println("I don't know about the color "+incolor+".");
}
}
cin.close();
}
}