Substring with concatenation of all Words

You is given a string, s, and a list of words, words, that is all of the same length. Find all starting indices of substring (s) in S that's a concatenation of each word in words exactly once and without any Intervening characters.

For example, given:
S"barfoothefoobarman"
Words["foo", "bar"]

You should return the indices: [0,9] .

Analysis:

The first way: add words to HashMap, and then starting from I, one by one to see if each word is in the hashmap, if present, then HashMap word in the corresponding count-1, if HashMap is empty, This means that all words have been acquired and the starting index is added to the list.

But it's not going to be the way. Wipe.

1  Public classSolution {2      PublicList<integer>findsubstring (String s, string[] words) {3list<integer> list =NewArraylist<>();4         if(s = =NULL|| S.length () = =0|| Words = =NULL|| Words.length = =0)5             returnlist;6 7         intLength = words[0].length ();8          for(inti =0; I < s.length (); i++) {9map<string, integer> map =Getmap (words);Ten             intindex =i; OneBoolean allcontained =true; A              while(!Map.isempty ()) { -                 if(Index + length >s.length ()) { -allcontained =false; the                      Break; -                 } -String sub = s.substring (index, Index +length); -                 if(Map.containskey (sub)) { +Map.put (sub, map.Get(sub)-1); -                     if(Map.Get(sub) = =0) { + Map.Remove (sub); A                     } atIndex + =length; -}Else { -allcontained =false; -                      Break; -                 } -             } in  -             if(allcontained) { to List.add (i); +             } -         } the         returnlist; *     } $ Panax Notoginseng     PrivateMap<string, integer>Getmap (string[] words) { -map<string, integer> map =NewHashmap<>(); the  +          for(String str:words) { AMap.put (str, map.getordefault (str,0) +1); the         } +         returnmap; -     } $}

Another way is to first a window to cover s, so that the word in the words is covered in the substring, then for the next, we need to move the left part of the window to the right word out, so that the right can continue to move. We also need to move the pointer on the left side of the window if the word we are adding to the right is already enough, or not in words. It is a pity that this method will not go through, no matter.

1  PublicList<integer>findsubstring (String s, string[] words) {2Map<string, integer> all =Getmap (words);3list<integer> list =NewArraylist<>();4         if(s = =NULL|| S.length () = =0|| Words = =NULL|| Words.length = =0)5             returnlist;6 7         intLength = words[0].length ();8          for(inti =0; i < length; i++) {9map<string, integer> map =Getmap (words);Ten             intindex =i; One             intstarting =i; A              while(Index + length <=s.length ()) { -String sub = s.substring (index, Index +length); -                 if(Map.containskey (sub)) { theMap.put (sub, map.Get(sub)-1); -                     if(Map.Get(sub) = =0) { - Map.Remove (sub); -                     } +                     //Find All the words in the HashMap -                     if(Map.isempty ()) { + List.add (starting); AMap.put (S.substring (starting, starting + length),1); atstarting = starting +length; -                     } -}Else { -                     if(All.containskey (sub)) { -Map.put (S.substring (starting, starting +length), -Map.getordefault (S.substring (starting, starting + length),0) +1); instarting = starting +length; -                         Continue; to}Else { +Map =Getmap (words); -Starting = index +length; the                     } *                 } $Index + =length;Panax Notoginseng             } -         } the         returnlist; +     } A  the     PrivateMap<string, integer>Getmap (string[] words) { +map<string, integer> map =NewHashmap<>(); -  $          for(String str:words) { $Map.put (str, map.getordefault (str,0) +1); -         } -         returnmap; the}

Substring with concatenation of all Words