Substring with Concatenation of All Words @LeetCode

package Level3;  
  
import java.util.ArrayList;  
import java.util.Hashtable;  
  
/** 
 * 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). 
 */  
public class S30 {  
  
    public static void main(String[] args) {  
        String S = "barfoothefoobarman";  
        String[] L = {"foo", "bar"};  
//      String S = "a";  
//      String[] L = {"a"};  
          
        System.out.println(findSubstring(S, L));  
    }  
      
    // 用暴力法压线通过，未来应该用类似KMP的方法降低复杂度！  
    public static ArrayList findSubstring(String S, String[] L) {  
          
        ArrayList ret = new ArrayList();  
          
        Hashtable ht = new Hashtable();  
        // 把L中的string全部添加入hashtable  
        for (String str : L) {  
            if(ht.containsKey(str)){  
                ht.put(str, ht.get(str)+1);  
            }else{  
                ht.put(str, 1);  
            }  
        }  
          
        int wordLen = L[0].length();  
        int numberOfWords = L.length;  
          
        // Brute force法比较  
        for(int i=0; i<=S.length()-numberOfWords*wordLen; i++){  
            // 每次要基于原hashtable新建一个hashtable  
            Hashtable





 table = new Hashtable(ht);  
            int cnt = 0;  
              
            // j每次都重复地做相同的工作  
            for(int j=i; j<=i+numberOfWords*wordLen-wordLen; j+=wordLen){  
                String substr = S.substring(j, j+wordLen);  
                if(table.containsKey(substr)){  
                    // 如果只用这一句会TLE  
//                  table.put(substr, table.get(substr)-1);  
                      
                    // 改成这样就能压线AC  
                    int times = table.get(substr);  
                    if(times == 1) table.remove(substr);  
                    else table.put(substr, times - 1);  
                      
                    cnt++;  
                }else{  
                    break;  
                }  
            }  
              
            if(cnt == numberOfWords){  
                ret.add(i);  
            }  
        }  
          
        return ret;  
    }  
  
}