Substring with Concatenation of All Words

 

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

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

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

第一遍：

 1 public class Solution {

 2      int elen = 0;

 3     public ArrayList<Integer> findSubstring(String S, String[] L) {

 4         // Note: The Solution object is instantiated only once and is reused by each test case.

 5         ArrayList<Integer> result = new ArrayList<Integer>();

 6         if(S == null || S.length() == 0) return result;

 7         int slen = S.length();

 8         int n = L.length;

 9         elen = L[0].length();

10         HashMap<String, Integer> hm = new HashMap<String, Integer>();

11         for(int i = 0; i < n; i ++){

12             if(hm.containsKey(L[i])) hm.put(L[i], hm.get(L[i]) + 1);

13             else                     hm.put(L[i], 1);

14         }

15         for(int i = 0; i <= slen - n * elen; i ++){

16             if(hm.containsKey(S.substring(i, i + elen)))

17                 if(checkOther(new HashMap<String, Integer>(hm), S, i))

18                     result.add(i);

19         }

20         return result;

21     }

22     public boolean checkOther(HashMap<String, Integer> hm, String s, int pos){

23         if(hm.size() == 0)  return true;

24         if(hm.containsKey(s.substring(pos, pos + elen))){

25             if(hm.get(s.substring(pos, pos + elen)) == 1)  hm.remove(s.substring(pos, pos + elen));

26             else                                           hm.put(s.substring(pos, pos + elen), hm.get(s.substring(pos, pos + elen)) - 1);

27             return checkOther(hm, s, pos + elen);

28         }

29         else return false;

30     }

31 }

 

第三遍：

Test case:

1. there can be duplicate in the word lists!!!

 1 public class Solution {

 2     int llen = 0;

 3     int lsize = 0;

 4     int slen = 0;

 5     List<Integer> result = null;

 6     public List<Integer> findSubstring(String S, String[] L) {

 7         result = new ArrayList<Integer> ();

 8         if(S == null || S.length() == 0 || L == null || L.length == 0 || L[0].length() == 0) return result;

 9         lsize = L.length;

10         llen = L[0].length();

11         slen = S.length();

12         HashMap<String, Integer> hs = new HashMap<String, Integer> ();

13         for(int i = 0; i < lsize; i ++){

14             if(!hs.containsKey(L[i]))

15                 hs.put(L[i], 1);

16             else

17                 hs.put(L[i], hs.get(L[i]) + 1);

18         }

19         for(int i = 0; i <= slen - lsize * llen; i ++)    

20             findString(S, i, new HashMap<String, Integer> (hs));

21         return result;

22     }

23     

24     public void findString(String s, int pos, HashMap<String, Integer> map){

25         if(map.size() == 0) result.add(pos - llen * lsize);

26         if(pos + llen <= slen && map.containsKey(s.substring(pos, pos + llen))){

27             if(map.get(s.substring(pos, pos + llen)) > 1)

28                 map.put(s.substring(pos, pos + llen), map.get(s.substring(pos, pos + llen)) - 1);

29             else

30                 map.remove(s.substring(pos, pos + llen));

31             findString(s, pos + llen, map);

32         }

33     }

34 }