[LeetCode] 527. Word Abbreviation 单词缩写

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.

Java:

    public List<String> wordsAbbreviation(List<String> dict) {
        int len=dict.size();
        String[] ans=new String[len];
        int[] prefix=new int[len];
        for (int i=0;i<len;i++) {
            prefix[i]=1;
            ans[i]=makeAbbr(dict.get(i), 1); // make abbreviation for each string
        }
        for (int i=0;i<len;i++) {
            while (true) {
                HashSet<Integer> set=new HashSet<>();
                for (int j=i+1;j<len;j++) {
                    if (ans[j].equals(ans[i])) set.add(j); // check all strings with the same abbreviation
                }
                if (set.isEmpty()) break;
                set.add(i);
                for (int k: set)
                    ans[k]=makeAbbr(dict.get(k), ++prefix[k]); // increase the prefix
            }
        }
        return Arrays.asList(ans);
    }

    private String makeAbbr(String s, int k) {
        if (k>=s.length()-2) return s;
        StringBuilder builder=new StringBuilder();
        builder.append(s.substring(0, k));
        builder.append(s.length()-1-k);
        builder.append(s.charAt(s.length()-1));
        return builder.toString();
    }　　

Java:

public List<String> wordsAbbreviation(List<String> dict) {
  String ans[] = new String[dict.size()];
  abbreviate(ans, dict, IntStream.range(0, ans.length).boxed().collect(Collectors.toList()), 1);
  return Arrays.asList(ans);
}

private void abbreviate(String[] ans, List<String> dict, List<Integer> idxs, int k) {
  Map<String, List<Integer>> map = new HashMap<>();
  idxs.stream().forEach(idx -> map.computeIfAbsent(getAbbr(dict.get(idx), k), key -> new ArrayList<>()).add(idx));
  for (Entry<String, List<Integer>> entry : map.entrySet())
    if (entry.getValue().size() == 1) ans[entry.getValue().get(0)] = entry.getKey();
    else abbreviate(ans, dict, entry.getValue(), k + 1);
}

private String getAbbr(String s, int k) {
  return s.length() - k < 3 ? s : s.substring(0, k) + (s.length() - 1 - k) + s.charAt(s.length() - 1);
}　　

Python:

class Solution(object):
    def wordsAbbreviation(self, dict):
        """
        :type dict: List[str]
        :rtype: List[str]
        """
        def isUnique(prefix, words):
            return sum(word.startswith(prefix) for word in words) == 1

        def toAbbr(prefix, word):
            abbr = prefix + str(len(word) - 1 - len(prefix)) + word[-1]
            return abbr if len(abbr) < len(word) else word

        abbr_to_word = collections.defaultdict(set)
        word_to_abbr = {}

        for word in dict:
            prefix = word[:1]
            abbr_to_word[toAbbr(prefix, word)].add(word)

        for abbr, conflicts in abbr_to_word.iteritems():
            if len(conflicts) > 1:
                for word in conflicts:
                    for i in xrange(2, len(word)):
                        prefix = word[:i]
                        if isUnique(prefix, conflicts):
                            word_to_abbr[word] = toAbbr(prefix, word)
                            break
            else:
                word_to_abbr[conflicts.pop()] = abbr

        return [word_to_abbr[word] for word in dict]　

Python:

class Solution(object):
    def abbr(self, word, size):
        if len(word) - size <= 3: return word
        return word[:size + 1] + str(len(word) - size - 2) + word[-1]

    def solve(self, dict, size):
        dlist = collections.defaultdict(list)
        for word in dict:
            dlist[self.abbr(word, size)].append(word)
        for abbr, wlist in dlist.iteritems():
            if len(wlist) == 1:
                self.dmap[wlist[0]] = abbr
            else:
                self.solve(wlist, size + 1)

    def wordsAbbreviation(self, dict):
        """
        :type dict: List[str]
        :rtype: List[str]
        """
        self.dmap = {}
        self.solve(dict, 0)
        return map(self.dmap.get, dict)　　

C++:

/*
Thought:
Originally, used a hashset to store all existing pattern. If checked word exist in dict hashset, then return false.
However, there is a case that: the word existed in the dict only for once, which is by accident the same as the checked work, then return true.
Therefore, we need to keep track of what word has been catagrize into pattern. SO, use a HashMap<String, ArrayList> instead.
Note: Dealing with char, integer, string. Be careful if char are turnning int integers.
*/
public class ValidWordAbbr {
	HashMap<String, ArrayList<String>> map;
    public ValidWordAbbr(String[] dict) {
        if (dict == null || dict.length == 0) {
        	return;
        }
        map = new HashMap<String, ArrayList<String>>();
        for (String s : dict) {
        	String str = "";
        	if (s.length() <= 2) {
        		str = s;
        	} else {
        		str += s.charAt(0) + (s.length() - 2 + "") + s.charAt(s.length() - 1);
        	}
        	if (!map.containsKey(str)) {
        		ArrayList<String> list = new ArrayList<String>();
 				list.add(s);
 				map.put(str, list);
        	} else {
        	    if (!map.get(str).contains(s)) {
        	       	map.get(str).add(s);
        	    }

        	}
        }
    }

    public boolean isUnique(String word) {
        if (map == null || map.size() == 0) {
            return true;
        }
        String str = "";
        if (word.length() <= 2) {
        	str =  word;
        } else {
        	str += word.charAt(0) + (word.length() - 2 + "") + word.charAt(word.length() - 1);
        }
        if (map.containsKey(str) && map.get(str).size() == 1 && map.get(str).get(0).equals(word)) {
        	return true;
        }
        return !map.containsKey(str);
    }
}

// Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa = new ValidWordAbbr(dictionary);
// vwa.isUnique("Word");
// vwa.isUnique("anotherWord");　

C++:

class Solution {
public:
    vector<string> wordsAbbreviation(vector<string>& dict) {
        int n = dict.size();
        vector<string> res(n);
        vector<int> pre(n, 1);
        for (int i = 0; i < n; ++i) {
            res[i] = abbreviate(dict[i], pre[i]);
        }
        for (int i = 0; i < n; ++i) {
            while (true) {
                set<int> s;
                for (int j = i + 1; j < n; ++j) {
                    if (res[j] == res[i]) s.insert(j);
                }
                if (s.empty()) break;
                s.insert(i);
                for (auto a : s) {
                    res[a] = abbreviate(dict[a], ++pre[a]);
                }
            }
        }
        return res;
    }
    string abbreviate(string s, int k) {
        return (k >= s.size() - 2) ? s : s.substr(0, k) + to_string(s.size() - k - 1) + s.back();
    }
};

类似题目：

[LeetCode] 288.Unique Word Abbreviation 独特的单词缩写

320. Generalized Abbreviation　

408.Valid Word Abbreviation

411. Minimum Unique Word Abbreviation

527. Word Abbreviation　

原文地址：https://www.cnblogs.com/lightwindy/p/9625356.html