Substring with concatenation of all Words, returns the position of strings that contain strings of string elements concatenated by strings

Problem Description: Given a character array words, and the string s, returns the position of a substring of all the character elements in the character array in the string, requiring all elements in the string array to exist only once in the string s.

Algorithmic analysis: This problem is similar to STRSTR. Just Strstr is a substring, and the question is a substring of elements in a string array, and the elements in the string array are unordered, but must all be included. All consider using the map collection. The key point is a few critical values, the string element repeated in s how to do, find a matching substring after how to do, there is a string element mismatch how to do.

Importjava.util.ArrayList;ImportJava.util.HashMap;Importjava.util.List; Public classSubstringwithconcatenationofallwords { PublicList<integer>findsubstring (String s, string[] words) {ArrayList<Integer> result =NewArraylist<integer>(); if(s==NULL|| S.length () ==0| | words==NULL|| Words.length==0){            returnresult; }              //frequency of wordshashmap<string, integer> map =NewHashmap<string, integer>();  for(String w:words) {if(Map.containskey (w)) {map.put (W, Map.get (w) )+1); }Else{map.put (W,1); }        }             intLen = words[0].length ();  for(intj=0; j<len; J + +) {HashMap<string, integer> currentmap =NewHashmap<string, integer>(); intstart = j;//starting subscript, because Len is the size of the window, so the starting subscript simply loops between the 0-len, just like modulo, over and over and over again.             intCount = 0;//record the number of strings that are satisfied                  for(intI=j; I<=s.length ()-len; i=i+Len) {String Sub= S.substring (i, i+Len); if(Map.containskey (sub)) {if(Currentmap.containskey (sub)) {Currentmap.put (Sub, currentmap.get ( sub))+1); }                    Else{currentmap.put (sub,1); } Count++; //whenever a substring is found to have elements in a repeating character array, the string is removed from the starting position. //In other words, the reverse operation is repeated until it contains only one repeating string.                      while(Currentmap.get (sub) >Map.get (sub)) {                        //four basic reverse operation. The window slide is repeated or the match succeeds. String left = s.substring (Start, start+Len); Currentmap.put (left, Currentmap.get)-1); Count--; Start= Start +Len; }                              if(count==words.length)//The string contains all the elements in the character array,{result.add (start);//Add starting position to result//four basic reverse operation. The window slides. Window value is LenString left = s.substring (Start, start+Len); Currentmap.put (left, Currentmap.get)-1); Count--; Start= Start +Len; }                }                Else{currentmap.clear (); Start= i+Len; Count= 0; }            }        }             returnresult; }}

The STRSTR algorithm is as follows:

 Public intstrStr (String haystack, string needle) {if(Haystack = =NULL|| Needle = =NULL)        {            return-1; }                 for(inti = 0; I <= (Haystack.length ()-needle.length ()); i++)         {            intm =i;  for(intj = 0; J < Needle.length (); J + +)             {                if(Needle.charat (j) = =Haystack.charat (M)) {m++; if((m-i) = =needle.length ()) {                        returni; }                }                 Else                 {                     Break; }            }        }        return-1; }

Substring with concatenation of all Words, returns the position of strings that contain strings of string elements concatenated by strings