LeetCode 527---Word Abbreviation

527. Word Abbreviation

Given an array of n distinct non-empty strings, you need to generate minimal possible abbreviations for every word following rules below.

1. Begin with the first character and then the number of characters abbreviated, which followed by the last character.
2. If there are any conflict, that is more than one words share the same abbreviation, a longer prefix is used instead of only the first character until making the map from word to abbreviation become unique. In other words, a final abbreviation cannot map to more than one original words.
3. If the abbreviation doesn't make the word shorter, then keep it as original.

Example:

Input: ["like", "god", "internal", "me", "internet", "interval", "intension", "face", "intrusion"]

Output: ["l2e","god","internal","me","i6t","interval","inte4n","f2e","intr4n"]

Note:

1. Both n and the length of each word will not exceed 400.
2. The length of each word is greater than 1.
3. The words consist of lowercase English letters only.
4. The return answers should be in the same order as the original array.

算法分析：

构造HashMap<String,ArrayList>abbre2Word,以每个字符串的Abbreviation做键，向该键下的ArrayList内添加映射到该键的Word。对于每个Abbreviation，如果其映射的ArrayList的size()为1，则该abbreviation为unique的，将该（Word，Abbreviation）添加到另一个 HashMap<String,String> word2Abbre,该映射以 Word 做键，以Abbreviation 做值；如果abbre2Word中的Abbreviation对应的ArrayList的 size() 大于1，则以此ArrayList做参数递归调用函数来重新生成Abbreviation，并且调用函数的时候传入 prefix 长度参数，该参数比上一次调用增加1。

Java算法实现：

public class Solution {
    public List<String> wordsAbbreviation(List<String> dict) {
		Map<String, String>map=new HashMap<>();
		WordMap2Abbreviation(map, 0, dict);
		List<String>result=new ArrayList<>();
		int size=dict.size();
		for(int i=0;i<size;i++){
			result.add(map.get(dict.get(i)));//调整map中Abbreviation的顺序，使result中的Abbreviation与dict中同一位置上的word相对应
		}
		return result;
	}
	
	public String getAbbreviation(String word,int fromIndex){//fromIndex表示从word的第几个字符开始生成缩写词
		int len=word.length();
		if(len-fromIndex<=3){//3个及以下的字符没有缩写的必要
			return word;
		}
		else{
			return word.substring(0, fromIndex+1)+String.valueOf(len-fromIndex-2)+word.charAt(len-1);
		}
	}
	
	public void WordMap2Abbreviation(Map<String, String>map,int fromIndex,List<String>dict){
		Map<String,ArrayList<String>>abbre2Word=new HashMap<>();//以abbreviation做键，value为Abbreviation相同的word组成的ArrayList<String>
		for(String word:dict){
			String abbre=getAbbreviation(word, fromIndex);
			if(abbre2Word.containsKey(abbre)){
				abbre2Word.get(abbre).add(word);
			}
			else{
				ArrayList<String>list=new ArrayList<>();
				list.add(word);
				abbre2Word.put(abbre, list);
			}
		}
		for(String abbre:abbre2Word.keySet()){
			ArrayList<String>words=abbre2Word.get(abbre);
			if(words.size()==1){//说明该Abbreviation是unique的
				map.put(words.get(0), abbre);
			}
			else{
				WordMap2Abbreviation(map, fromIndex+1, words);//对这些Abbreviation相同的word递归调用函数
			}
		}
	}
}