Leetcode Substring with concatenation of all Words (C,c++,java,python)

Problem:

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 wordsexactly once And without any intervening characters.

for example, given:
s :  " Barfoothefoobarman "
words :  [" foo "," Bar "]

You should return the indices: [0,9] .
(Order does not matter).

Solution: Because the length is fixed, fixed the position where each string starts, then searches backwards to see if each string exists, if it does not exist i++
Java source Code (658MS):

public class Solution {public list<integer> findsubstring (String s, string[] words) {Map<string,integ        Er> tmp,map=new hashmap<string,integer> ();        int Lens=s.length (), lenw=words[0].length (); for (int i=0;i<words.length;i++) {if (Map.containskey (words[i))) {Map.put (Words[i],map.get (Word            S[i]) +1);            }else{Map.put (words[i],1);        }} list<integer> res = new arraylist<integer> ();            for (int i=0;i<=lens-lenw*words.length;i++) {tmp=new hashmap<string,integer> ();            int j=0;                for (; j<words.length;j++) {int pos=i+j*lenw;                String sub=s.substring (POS,POS+LENW);                    if (Map.containskey (sub)) {int num=0;                    if (Tmp.containskey (sub)) Num=tmp.get (sub);                    if (Map.get (sub) < num+1) break;               else Tmp.put (sub,num+1); }else break;            } if (j>=words.length) {res.add (i);    }} return res; }}

C Language Source code (260MS): Sub Not Free will be MLE

typedef struct node{char* Word;    int times; struct node* Next;}    data; #define SIZE 1000int hash (char* word) {int i=0,h=0;    for (; word[i];i++) {h= (h*31+word[i])%size; } return h;}    int Insertmap (data** map,char* word,int lenw) {int h = hash (word);        if (map[h]==null) {map[h]= (data*) malloc (sizeof (data));        Map[h]->word= (char*) malloc (sizeof (char) * (lenw+1));        map[h]->times=1;strcpy (Map[h]->word,word);        map[h]->next=null;    return 1;        }else{data* P=map[h];                while (P->next!=null) {if (strcmp (P->word,word) ==0) {p->times++;            Return p->times;        } p=p->next;            } if (strcmp (P->word,word) ==0) {p->times++;        Return p->times;            }else{data* tmp= (data*) malloc (sizeof (data));            Tmp->word= (char*) malloc (sizeof (char) * (lenw+1));      tmp->times=1;strcpy (Tmp->word,word);      tmp->next=null;            p->next=tmp;        return 1;    }}}int Findmap (data** map,char* sub) {int h = hash (sub);    if (map[h]==null) {return-1;        }else{data* P=map[h];            while (P!=null) {if (strcmp (p->word,sub) ==0) {return p->times;        } p=p->next;    } return-1; }}char* substring (char* s,int start,int len) {char* sub= (char*) malloc (sizeof (char) * (len+1)); int i=0;for (; i<len;i++ ) {Sub[i]=s[i+start];} Sub[i]=0;return Sub;} int* findsubstring (char* s, char** words, int wordssize, int* returnsize) {int Lenw=strlen (words[0]), Lens=strlen (s), le    Ngth=wordssize;    int* res= (int*) malloc (sizeof (int) * (lens-lenw*length+1));    data** map= (data**) malloc (sizeof (data*) *size);    data** tmp= (data**) malloc (sizeof (data*) *size);    int I,j;for (i=0;i<size;i++) {map[i]=null;tmp[i]=null;}    for (i=0;i<length;i++) {insertmap (MAP,WORDS[I],LENW); }*returnsize=0;for (I=0;i<lens-lenw*length+1;i++) {for (j=0;j<size;j++) {if (tmp[j]!=null) {free (tmp[j]); tmp[j]=null;}} for (j=0;j<length;j++) {char* sub=substring (s,i+j*lenw,lenw), int mapnum=findmap (map,sub), if (mapnum==-1) break; int Num=insertmap (TMP,SUB,LENW), if (Mapnum < num) Break;free (sub);} if (j>=length) res[(*returnsize) ++]=i;} for (i=0;i<size;i++) if (map[i]!=null) free (map[i]), free (map); return res;}

C + + source code (704MS):

Class Solution {public:    vector<int> findsubstring (string s, vector<string>& words) {        int lens= S.size (), Lenw=words[0].size (), length=words.size ();        Map<string,int> strmap,tmp;        for (int i=0;i<length;i++) {            strmap[words[i]]++;        }        vector<int> Res;        for (int i=0;i<lens-lenw*length+1;i++) {            tmp.clear ();            int j=0;            for (j=0;j<length;j++) {                string sub=s.substr (I+J*LENW,LENW);                if (Strmap.find (sub) = = Strmap.end ()) break;                tmp[sub]++;                if (Strmap[sub] < tmp[sub]) break;            }            if (j>=length) res.push_back (i);        }        return res;    }};

Python source code (706MS):

Class solution:    # @param {string} s    # @param {string[]} words    # @return {integer[]}    def findsubstring ( Self, S, words):        Lens=len (s); Lenw=len (Words[0]); Length=len (words)        map={};res=[] for        i in range (length):            if words[i] in map:map[words[i]]+=1            else:map[words[i]]=1            if not words[i] in S:return res for        I in range (lens-lenw*length+1):            tmp={};j=0;flag=true for            J in range (length):                pos=i+j*lenw                Sub=s[pos:pos +LENW]                if sub in map:                    num=0                    if sub in tmp:num=tmp[sub]                    if map[sub] < Num+1:flag=false;break                    tmp[sub]=num+1                else:flag=false;break            if Flag:res.append (i)        return res

Leetcode Substring with concatenation of all Words (C,c++,java,python)