[Lintcode Easy] Longest Words

Longest Words

Given A dictionary, find all of the longest words in the dictionary.

Example

Given

{  "dog",  "google",  "facebook",  "internationalization",  "blabla"}

The longest words is ["internationalization"] .

Given

{  "like",  "love",  "hate",  "yes"}

The longest words is ["like", "love", "hate"] .

Challenge

It's easy-to-solve it in the passes, can do it in one pass?

First,tracking the array to record every word ' s length, and find out the max length

The second loop is for find out specific words.

Keep in mind how to initiate the int array.

classSolution {/**     * @paramDictionary:an Array of strings *@return: An ArrayList of strings*/ArrayList<String>longestwords (string[] dictionary) {//Write your code hereArrayList<String> newstring=NewArraylist<string>(); intcount[]=NULL; intm=dictionary.length; Count=New int[M]; intMax=0;  for(inti=0;i<dictionary.length;i++)        {            intn=dictionary[i].length (); Count[i]=N; Max=Math.max (N,max); }                 for(inti=0;i<dictionary.length;i++)        {            if(count[i]==max)            {Newstring.add (dictionary[i]); }        }                returnnewstring; }

If we use Hashtable, we can does it in one pass

Following is the code,

classSolution {/**     * @paramDictionary:an Array of strings *@return: An ArrayList of strings*/ArrayList<String>longestwords (string[] dictionary) {//Write your code hereHashMap<Integer,ArrayList<String>> map=NewHashmap<integer,arraylist<string>>(); intMax=0;  for(inti=0;i<dictionary.length;i++)        {           if(Map.containskey (Dictionary[i].length ())) {Map.get (Dictionary[i].length ()). Add (Dictionary[i]           ); }            Else{ArrayList<String> arr=NewArraylist<string>();               Arr.add (Dictionary[i]);           Map.put (Dictionary[i].length (), arr); } Max=Math.max (Dictionary[i].length (), Max); }        returnMap.get (max); }};

[Lintcode Easy] Longest Words