leetcode--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).

 

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public class Solution{	
	public static List<Integer> findSubstring(String S, String[] L){
		ArrayList<Integer> startPoints = new ArrayList<Integer>();
		if(L.length != 0){
			if(S== null || S.length() == 0)
				return startPoints;
			else{
				int lengthOfArray = L.length;
				int lengthOfWord = L[0].length();
				int lengthOfString = S.length();
				if(lengthOfString < lengthOfArray * lengthOfWord)
					return startPoints;
				else{
					//put the string[] into a dict
					Map<String, Integer> dict = new HashMap<String, Integer>();
					for(int j = 0; j < lengthOfArray; ++j){
						if(dict.containsKey(L[j])){
							int value = dict.get(L[j]) + 1;
							dict.put(L[j], value);
						}
						else
							dict.put(L[j], 1);
					}
					
					for(int i = 0; i < lengthOfString - lengthOfWord * lengthOfArray + 1; ++i){
						//check each substring with length lengthOfWord * lengthOfArray starting from index i
						Map<String, Integer> check = new HashMap<String, Integer>();
						boolean canBeStart = true;
						for(int j = 0; j < lengthOfArray; ++j){
							String inChecked = S.substring(i + j * lengthOfWord, i + (j + 1)*lengthOfWord);
							if(dict.containsKey(inChecked)){
								if(check.containsKey(inChecked)){
									int mult = check.get(inChecked) + 1;
									if(mult > dict.get(inChecked)){
										canBeStart = false;
										break;
									}
									else
										check.put(inChecked, mult);										
								}
								else
									check.put(inChecked, 1);
							}
							else{
								canBeStart = false;
								break;
							}
						}						
						if(canBeStart)
							startPoints.add(i);						
					}
				}
			}			
		}
		return startPoints;
	}
}