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
Solution：hashmap
思路：前提遍歷存好(abr_key -> str)的map品腹，如果有重復(fù)当宴，將str置為""使之無(wú)效這樣要判斷的同key的str和它一定不相等，return false。
Time Complexity: 初始化構(gòu)造O(N) isUnique(): O(1) Space Complexity: O(N)
Solution Code：
public class ValidWordAbbr {
HashMap<String, String> map;
public ValidWordAbbr(String[] dictionary) {
map = new HashMap<String, String>();
for(String str: dictionary){
String key = getKey(str);
// If there is more than one string belong to the same key
// then the key will be invalid, we set the value to ""
if(map.containsKey(key)){
if(!map.get(key).equals(str)){
map.put(key, "");
}
}
else{
map.put(key, str);
}
}
}
public boolean isUnique(String word) {
return !map.containsKey(getKey(word)) || map.get(getKey(word)).equals(word);
}
String getKey(String str){
if(str.length() <= 2) return str;
return str.charAt(0)+Integer.toString(str.length() - 2)+str.charAt(str.length() - 1);
}
}
