題目：

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 --> d1g1 1 11---5----0----5--8 c) i|nternationalizatio|n --> i18n11---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

鏈接：?http://leetcode.com/problems/unique-word-abbreviation/

題解：

新題的題目真是越來越長了。 這道題是給定一個數(shù)組Dictionary， 求輸入字符串是否有unique的abbreviation在Dictionary中。我們選擇用Map<String, Set<String>>來做我們存儲數(shù)據(jù)的數(shù)據(jù)結構，然后按照題意做就可以了，還需要判斷一些邊界條件，比如縮寫不在map中直接返回true之類的。 以后一定要牢記，選定了好的數(shù)據(jù)結構，編寫程序就會容易很多。

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:

跟一刷的方法一樣。就是跟Anagram一樣，用Map<String, Set<String>>來存，使用一個新的方法getAbbr先求出abbr作為key，然后把單詞加入到key的value里。 ?Discuss里面還有很多很好的方法，用map<String, String>之類的，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?

轉載于:https://www.cnblogs.com/yrbbest/p/5040715.html

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