十进制数转罗马数字 – Switch笨方法解题

小半 • 2022年10月28日下午7:12 • 其他分类 • 阅读 224

数字转罗马数字工具类

class NumberChange {

	public static String reverseStr(String str) {
		StringBuilder sb = new StringBuilder();
		int index = str.length() - 1;
		for (int i = index; i >= 0; i--) {
			sb.append(str.substring(i, i + 1));
		}
		return sb.toString();
	}

	public static String currentLocationInvert( String number , Integer zeroCount) {
		
		Integer num = Integer.parseInt(number);
		StringBuilder sb = new StringBuilder();
		switch(zeroCount) {
			case 0 : {
				
				 if(num<4) {
						for(int i = 0; i < num; i++) {
							sb.append("I");
						}
				}else if(num == 4) {
						sb.append("IV");
				}else if(num == 5) {
						sb.append("V");
				}else if(num == 9) {
						sb.append("IX");
				}else {
						sb.append("V");
						for(int i = 0; i<num-5; i++) {
							sb.append("I");
						}
					}
				 break;
			}
				
			case 1 :  {
				
				 if(num<4) {
						for(int i = 0; i < num; i++) {
							sb.append("X");
						}
				}else if(num == 4) {
						sb.append("XL");
				}else if(num == 5) {
						sb.append("L");
				}else if(num == 9) {
						sb.append("XC");
				}else {
						sb.append("L");
						for(int i = 0; i<num-5; i++) {
							sb.append("X");
						}
					}
				 break;
			}
			case 2 :   {
				
				 if(num<4) {
						for(int i = 0; i < num; i++) {
							sb.append("C");
						}
				}else if(num == 4) {
						sb.append("CD");
				}else if(num == 5) {
						sb.append("D");
				}else if(num == 9) {
						sb.append("CM");
				}else {
						sb.append("D");
						for(int i = 0; i<num-5; i++) {
							sb.append("C");
						}
					}
				 break;
			}
			case 3  : {
				for(int i = 0; i < num; i++) {
					sb.append("M");
				}
			}
		}

	return sb.toString();

	}

	public static String invertToRoman(Integer number) {
		
		// 1. 只转 1-3999范围的数字
		if(number<1 || number > 3999) {
			throw new RuntimeException("请输入在【1~3999】范围内的数字 - 小老弟别搞事");
		}
		
		// 2. 将数字倒序
		String strNum = reverseStr(number.toString());
		
		// 3. 定义一个字符输出对象
		StringBuilder sb = new StringBuilder();
		
		// 4. 遍历输出 - 打印
		for(int i = strNum.length()-1; i>=0; i--) {
			String num = strNum.substring(i,i+1);
			String roma = currentLocationInvert(num, i);
			sb.append(roma);
		}
		
		// 5. 返回结果
		return sb.toString();
	}

}

测试代码

public static void main(String[] args) {
    System.out.println(NumberChange.invertToRoman(1994));
    System.out.println(NumberChange.invertToRoman(58));
    System.out.println(NumberChange.invertToRoman(9));
    System.out.println(NumberChange.invertToRoman(4));
    System.out.println(NumberChange.invertToRoman(3));
}