Word Break II

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

参考：http://www.tuicool.com/articles/ZVJJRrF

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Set;

public class Solution {
    public List<String> wordBreak(String s, Set<String> dict) {
        List<String> dp[] = new ArrayList[s.length() + 1];
        dp[0] = new ArrayList<String>();
        
        for(int i = 0; i < s.length(); i++){
            if(dp[i] == null)
                continue;
            for(String word : dict){
                int length = word.length();
                int end = i + length;
                if(end > s.length())
                    continue;
                if(s.substring(i, end).equals(word)){
                    if(dp[end] == null)
                        dp[end] = new ArrayList<String>();
                    dp[end].add(word);
                }//if
            }//for
        }//for
        List<String> result = new LinkedList<String>();
        if(dp[s.length()] == null)
            return result;
        List<String> temp = new ArrayList<String>();
        
        dfsStringList(dp, result, s.length(), temp);
        
        return result;
    }
    
    private void dfsStringList(List<String> dp[], List<String> ans, 
            int end, List<String> temp){
        if(end <= 0){
            String strAns = temp.get(temp.size() - 1);
            for(int i = temp.size() - 2; i >= 0; i--)
                strAns += " " + temp.get(i);
            ans.add(strAns);
            return;
        }//if
        for(String str : dp[end]){
            temp.add(str);
            dfsStringList(dp, ans, end - str.length(), temp);
            temp.remove(temp.size() - 1);
        }
    }
}