主函数的剪枝,判断单词是否由数组中的单词组成的函数的记忆化。
public final String longestWord(String[] words) { Arrays.sort(words, (o1, o2) -> { if (o1.length() == o2.length()) return o1.compareTo(o2); else return Integer.compare(o2.length(), o1.length()); }); Set<String> set = new HashSet<String>(); for (String word : words) set.add(word); for (String word : words) { set.remove(word); if (isCompose(word, word.length(), set, new int[word.length() + 1])) return word; } return ""; } private final boolean isCompose(String target, int point, Set<String> set, int[] cache) { if (point == 0) return true; if (cache[point] != 0) return cache[point] == 1; for (int i = point - 1; i >= 0; i--) { String last = target.substring(i, point); if (set.contains(last) && isCompose(target, i, set, cache)) { cache[point] = 1; break; } else cache[point] = -1; } return cache[point] == 1; }