[codewars][Java] 翻译摩斯码

In this kata you have to write a simple Morse code decoder. While the Morse code is now mostly superceded by voice and digital data communication channels, it still has its use in some applications around the world.

The Morse code encodes every character as a sequence of "dots" and "dashes". For example, the letter A is coded as ·−, letter Q is coded as −−·−, and digit 1 is coded as ·−−−−. The Morse code is case-insensitive, traditionally capital letters are used. When the message is written in Morse code, a single space is used to separate the character codes and 3 spaces are used to separate words. For example, the message HEY JUDE in Morse code is ···· · −·−−   ·−−− ··− −·· ·.

NOTE: Extra spaces before or after the code have no meaning and should be ignored.

In addition to letters, digits and some punctuation, there are some special service codes, the most notorious of those is the international distress signal SOS (that was first issued by Titanic), that is coded as ···−−−···. These special codes are treated as single special characters, and usually are transmitted as separate words.

Your task is to implement a function that would take the morse code as input and return a decoded human-readable string.

For example:

MorseCodeDecoder.decode(".... . -.--   .--- ..- -.. .")
//should return "HEY JUDE"

我的code：

import java.util.ArrayList;

public class MorseCodeDecoder {
    public static String decode(String morseCode) {
        // your brilliant code here, remember that you can access the preloaded Morse code table through MorseCode.get(code)
    	 String[] list = morseCode.split(" ");   //用空格分隔，并存入一个数组中
		 ArrayList<String> arr = new ArrayList<>();
		 int space = 0;
         int count =0;
		 String s = "";

		for(int i =0 ; i<list.length ; i++)
		 {
			 arr.add(MorseCode.get(list[i]));
		 }
		for(int k=0; k<arr.size(); k++)
		{
			
			if(arr.get(k) == null)
			{
				space ++;
				
			}
			else    //第k个元素不等于null
			{
				if(space ==0)
				{
					s += arr.get(k);
					count++;
				}
				else
				{
					if(count ==0)    //前面的元素为空
					{
						s += arr.get(k);
						space = 0;
						count ++;
					}
					else
					{
						s += " "+arr.get(k);
						space = 0;	
					}
				}
						
			}
			
		}
		
		 return s;
	 }

}


分几种情况进行讨论：
1. 链表中元素 = null
   1> 将 space ++；， 并且不做任何输出
2. 链表中元素非空
   1> space = 0， 说明当前元素之前没有null元素，可以直接输出；如 [E,null,null,E]
      s = s+ arr.get(k);
   2> space >0, 说明当前元素前为null, 此时需要判断null元素是出现在链表开头，如： 
      [null,null,E,null,null,E]; 还是中间位置,如： [E,null,null,E]
      ~ 如果null出现在开头，如[null,null,E,null,null,E],则标志变量count=0；
      ~ 如果null出现在中间，如[E,null,null,E]， 则标志变量count =1;

Clever code：

public class MorseCodeDecoder {
    public static String decode(String morseCode) {
      String result = "";
      for(String word : morseCode.trim().split("   ")) {
        for(String letter : word.split("\\s+")) {
          result += MorseCode.get(letter);
        }
        result += ' ';
      }
      return result.trim();
    }
}

// .trim() 方法用于删除字符串两端的字符
// "\\s+" 为正则表达式，"\"为转义字符，"\s+" 代表多个空白字符，可以是空格，回车，换行，换页，制表符等。这个正则表达式经常用到