leetcode WordBreakII

import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;

public class WordBreakII {
  /**
   * Given a non-empty string s and a dictionary wordDict containing a list of non-empty words,
   * add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.
   *
   * Note:
   *
   * The same word in the dictionary may be reused multiple times in the segmentation.
   * You may assume the dictionary does not contain duplicate words.
   * Example 1:
   *
   * Input:
   * s = "catsanddog"
   * wordDict = ["cat", "cats", "and", "sand", "dog"]
   * Output:
   * [
   *   "cats and dog",
   *   "cat sand dog"
   * ]
   * Example 2:
   *
   * Input:
   * s = "pineapplepenapple"
   * wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
   * Output:
   * [
   *   "pine apple pen apple",
   *   "pineapple pen apple",
   *   "pine applepen apple"
   * ]
   * Explanation: Note that you are allowed to reuse a dictionary word.
   * Example 3:
   *
   * Input:
   * s = "catsandog"
   * wordDict = ["cats", "dog", "sand", "and", "cat"]
   * Output:
   * []
   */
  public List<String> wordBreak(String s, List<String> wordDict) {
    HashMap<String, List<String>> map = new HashMap<>();
    bfs( s, map, wordDict);
    return map.get(s);
  }

  public List<String> bfs(String s, HashMap<String, List<String>> map, List<String> wordDick) {
    if (map.containsKey(s)) {
      return map.get(s);
    }

    List<String> t = new ArrayList<>();

    if (s.isEmpty()) {
      t.add("");
    }

    for (String word: wordDick) {
      if (s.startsWith(word)) {
        List<String> left = bfs( s.substring(word.length()), map, wordDick);
        if (! left.isEmpty()) {
          for (String l: left) {
            t.add(word + (l.isEmpty()?"":" ") + l);
          }
        }
      }
    }
    map.put(s, t);
    return t;
  }
}