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).
使用了 HashMap来降低时间复杂度， 整个的算法很简单。
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])) 13 hm.put(L[i], hm.get(L[i]) + 1); 14 else 15 hm.put(L[i], 1); 16 } 17 for(int i = 0; i <= slen - n * elen; i ++){ 18 if(hm.containsKey(S.substring(i, i + elen))) 19 if(checkOther(new HashMap<String, Integer>(hm), S, i)) 20 result.add(i); 21 } 22 return result; 23 } 24 public boolean checkOther(HashMap<String, Integer> hm, String s, int pos){ 25 if(hm.size() == 0) return true; 26 if(hm.containsKey(s.substring(pos, pos + elen))){ 27 if(hm.get(s.substring(pos, pos + elen)) == 1) 28 hm.remove(s.substring(pos, pos + elen)); 29 else 30 hm.put(s.substring(pos, pos + elen), hm.get(s.substring(pos, pos + elen)) - 1); 31 return checkOther(hm, s, pos + elen); 32 } 33 else return false; 34 } 35 }