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 }

 

posted on 2014-08-12 10:48  Step-BY-Step  阅读(139)  评论(0)  编辑  收藏  举报

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