给定一个字符串 String s = "leetcode"
dict = ["leet", "code"].

查看一下是够是字典中的词语组成。假设是返回true，否则返回false。

下边提供3种思路

1.动态算法

import java.util.HashSet;
import java.util.Set;

public class WordBreak1 {
	public static void main(String[] args) {
		//"["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]
		//String s="aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab";
		String s ="LeetCodea";
		Set<String> dict = new HashSet<String>();
		dict.add("Leet");
		dict.add("Code");
		dict.add("a");
		System.out.println(wordBreak(s,dict));
	}
	public static boolean wordBreak(String s, Set<String> dict) {
		boolean[] t = new boolean[s.length() + 1];
		t[0] = true; // set first to be true, why?
		// Because we need initial state

		for (int i = 0; i < s.length(); i++) {
			// should continue from match position
			if (!t[i])
				continue;
			for (String a : dict) {
				int len = a.length();
				int end = i + len;
				if (end > s.length())
					continue;

				if (t[end])
					continue;

				if (s.substring(i, end).equals(a)) {
					t[end] = true;
				}
			}
		}
		return t[s.length()];
	}
}

2.普通算法(1)

import java.util.Set;

public class WorkBreak2 {
	public boolean wordBreak(String s, Set<String> dict) {
		return wordBreakHelper(s, dict, 0);
	}

	public boolean wordBreakHelper(String s, Set<String> dict, int start) {
		if (start == s.length())
			return true;

		for (String a : dict) {
			int len = a.length();
			int end = start + len;

			// end index should be <= string length
			if (end > s.length())
				continue;

			if (s.substring(start, start + len).equals(a))
				if (wordBreakHelper(s, dict, start + len))
					return true;
		}
		return false;
	}
}

3.普通算法(2)

import java.util.Set;

public class WordBreak3 {

	public static boolean wordBreak(String s, Set<String> dict) {
		// input validation
		// Base case
		if (dict.contains(s))
			return true;
		else {
			for (int i = 0; i < s.length(); i++) {
				String sstr = s.substring(0, i);
				if (dict.contains(sstr))
					return wordBreak(s.substring(i), dict);
			}
		}
		return false;
	}
}

可是以上的算法有一个问题，就是遇到这样的情况。INPUT: "programcreek", ["programcree","program","creek"]. 无能为力。

大家讨论下吧？