用了两个办法：

（1）试着写了个前缀树，352 ms。

  其中最大的trick是要用一个数组 boolean[] searchFlag来记录从s[i]开始的子串(s[i...n-1])是否被搜索过（s[i...n-1]能否由dict的单词组成)。如果被搜索过那就不用再搜索了（因为搜索失败了，如果成功的话程序早已结束）。

import java.util.*;

public class Solution {

	class TrieNode {
		char ch;
		ArrayList<TrieNode> children = new ArrayList<TrieNode>();

		TrieNode(char c) {
			ch = c;
		}

		TrieNode() {
			ch = '\0';
		}
	}

	public void insert(String word, int ind, TrieNode node) {
		if (ind == word.length()) {
			TrieNode nNode = new TrieNode('\0');
			node.children.add(nNode);
			return;
		}

		int i;
		for (i = 0; i < node.children.size(); i++) {
			if (node.children.get(i).ch == word.charAt(ind)) {
				insert(word, ind + 1, node.children.get(i));
				break;
			}
		}

		if (i == node.children.size()) {
			TrieNode nNode = new TrieNode(word.charAt(ind));
			node.children.add(nNode);
			insert(word, ind + 1, nNode);
		}
	}

	public boolean search(String s, int ind, TrieNode root, TrieNode node, boolean[] searchFlag) {
		if (ind == s.length()) {
			for (int i = 0; i < node.children.size(); i++)
				if (node.children.get(i).ch == '\0')
					return true;
			return false;
		}
		
		if(node == root && searchFlag[ind] == true)//the sub string[i...n-1] has been searched before
			return false;

		int i;
		boolean flag = false;
		for (i = 0; i < node.children.size(); i++) {
			if (node.children.get(i).ch == '\0') {
				// end of word, try to search a new word again
					flag = search(s, ind, root, root, searchFlag);
					searchFlag[ind] =true;
					
			} else if (node.children.get(i).ch == s.charAt(ind)) {
				// go on iterate the word
				flag = search(s, ind + 1, root, node.children.get(i), searchFlag);
			}

			if (flag == true)
				return true;
		}
		

		return false;
	}

	public boolean wordBreak(String s, Set<String> dict) {

		TrieNode root = new TrieNode();

		for (String word : dict) {
			insert(word, 0, root);
		}

		boolean[] searchFlag = new boolean[s.length()];
		for(int i=0;i<s.length();i++)
				searchFlag[i] = false;
		
		boolean flag = search(s, 0, root, root, searchFlag);

		return flag;
	}

	public static void main(String args[]) {
		Solution sol = new Solution();
		String s = "abcd";
		Set<String> dict = new HashSet<String>(Arrays.asList("a","abc","b","cd"));

		System.out.println(sol.wordBreak(s, dict));
	}
}

（2）不用前缀树，直接在set里面查找，因为数据量少的缘故变得更加快了。336ms

public class Solution {

	public boolean search(String s, int beg, Set<String> dict, boolean[] searchFlag)
	{
		if(beg == s.length())
			return true;
		
		if(searchFlag[beg] == true)
			return false;
		
		boolean flag = false;
		for(int e = beg;e<=s.length();e++)
		{
			String subStr = s.substring(beg,e);
			if(dict.contains(subStr))
			{
				flag = search(s,e,dict,searchFlag);
				if(flag == true)
					return true;
				
				searchFlag[e] = true;
			}
		}
		
		return false;
	}

	public boolean wordBreak(String s, Set<String> dict) {
		boolean[] searchFlag = new boolean[s.length()+1];
		
		for(int i=0;i<s.length()-1;i++)
			searchFlag[i] = false;
		
		boolean flag = search(s,0,dict, searchFlag);
		
		return flag;
	}

}

Leetcode - WordBreak