題目：
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
Example 1:
Input:
s = "barfoothefoobarman",
words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoo" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.
Example 2:
Input:
s = "wordgoodgoodgoodbestword",
words = ["word","good","best","word"]
Output: []
思路：
遍歷字符串s，每次取出和words一樣長的字串，分割成數組subs，比較數組subs和words，若相同（不考慮順序），則找到一個index放入結果集中。
代碼：
public List<Integer> findSubstring(String s, String[] words) {
List<Integer> res = new LinkedList<>();
if(words == null || words.length == 0) return res;
int wordLength = words[0].length();
for(int i = 0; i <= s.length() - wordLength*words.length; ++i){
String[] subs = new String[words.length];
int k = i;
for(int j = 0; j < words.length; ++j) {
subs[j] = s.substring(k, k + wordLength);
k += wordLength;
}
if(compareTwoArrays(subs, words)){
res.add(i);
}
}
return res;
}
public boolean compareTwoArrays(String[] subs, String[] words){
HashMap<String, Integer> map = new HashMap<>();
for(int i = 0; i < words.length; ++i){
if(map.getOrDefault(words[i], 0) == 0){
map.put(words[i],1);
}else{
map.put(words[i], map.get(words[i])+1);
}
}
for(int i = 0; i < subs.length; ++i){
if(map.getOrDefault(subs[i],0) == 0)
return false;
else
map.put(subs[i], map.get(subs[i])-1);
}
for(int i = 0; i < words.length; ++i){
if(map.getOrDefault(words[i], 0) != 0){
return false;
}
}
return true;
}