288. Unique Word Abbreviation

题目：

An abbreviation of a word follows the form <first letter><number><last letter>. Below are some examples of word abbreviations:

a) it                      --> it    (no abbreviation)     1
b) d|o|g                   --> d1g              1    1  1
     1---5----0----5--8
c) i|nternationalizatio|n  --> i18n              1
     1---5----0
d) l|ocalizatio|n          --> l10n

Assume you have a dictionary and given a word, find whether its abbreviation is unique in the dictionary. A word’s abbreviation is unique if no other word from the dictionary has the same abbreviation.

Example:

Given dictionary = [ "deer", "door", "cake", "card" ]isUnique("dear") -> false
isUnique("cart") -> true
isUnique("cane") -> false
isUnique("make") -> true

链接： http://leetcode.com/problems/unique-word-abbreviation/

题解：

新题的题目真是越来越长了。 这道题是给定一个数组Dictionary， 求输入字符串是否有unique的abbreviation在Dictionary中。我们选择用Map<String, Set<String>>来做我们存储数据的数据结构，然后按照题意做就可以了，还需要判断一些边界条件，比如缩写不在map中直接返回true之类的。 以后一定要牢记，选定了好的数据结构，编写程序就会容易很多。

Time Complexity – O(n * L)， Space Complexity – O(n * L)。

public class ValidWordAbbr {
    private Map<String, HashSet<String>> map;    public ValidWordAbbr(String[] dictionary) {
        this.map = new HashMap<>();
        for(int i = 0; i < dictionary.length; i++) {
            String abbr = getAbbr(dictionary[i]);
            if(!map.containsKey(abbr)) {
                HashSet<String> set = new HashSet<>();
                set.add(dictionary[i]);
                map.put(abbr, set);
            } else {
                if(!map.get(abbr).contains(dictionary[i])) {
                    map.get(abbr).add(dictionary[i]);
                }
            }
        }
    }    public boolean isUnique(String word) {
        if(map.size() == 0 || word.length() < 3) {
            return true;
        }
        String abbr = getAbbr(word);
        if(!map.containsKey(abbr) || (map.get(abbr).contains(word) && map.get(abbr).size() == 1)) {
            return true;
        } else {
            return false;
        }
    }    private String getAbbr(String s) {
        if(s.length() < 3) {
            return s;
        } else {
            return s.substring(0, 1) + String.valueOf(s.length() - 2) + s.substring(s.length() - 1);
        }
    }
}// Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa = new ValidWordAbbr(dictionary);
// vwa.isUnique("Word");
// vwa.isUnique("anotherWord");

二刷:

跟一刷的方法一样。就是跟Anagram一样，用Map<String, Set<String>>来存，使用一个新的方法getAbbr先求出abbr作为key，然后把单词加入到key的value里。  Discuss里面还有很多很好的方法，用map<String, String>之类的，三刷要好好研究。

Java:

Time Complexity – O(n * L)， Space Complexity – O(n * L)。

public class ValidWordAbbr {
    Map<String, Set<String>> map;
    public ValidWordAbbr(String[] dictionary) {
        map = new HashMap<>();
        for (String s : dictionary) {
            String abbr = getAbbr(s);
            if (!map.containsKey(abbr)) {
                map.put(abbr, new HashSet<String>());
            }
            map.get(abbr).add(s);
        }
    }    public boolean isUnique(String word) {
        String abbr = getAbbr(word);
        if (!map.containsKey(abbr) || (map.get(abbr).contains(word) && map.get(abbr).size() == 1)) {
            return true;
        }
        return false;
    }    private String getAbbr(String s) {
        if (s.length() < 3) {
            return s;
        }
        int len = s.length();
        return s.substring(0, 1) + (len - 2) + s.substring(len - 1);
    }
}// Your ValidWordAbbr object will be instantiated and called as such:
// ValidWordAbbr vwa = new ValidWordAbbr(dictionary);
// vwa.isUnique("Word");
// vwa.isUnique("anotherWord");

Reference:

https://leetcode.com/discuss/62842/a-simple-java-solution-using-map-string-string

https://leetcode.com/discuss/61658/share-my-java-solution

https://leetcode.com/discuss/71652/java-solution-with-hashmap-string-string-beats-submissions