这道题相似  Word Break 推断能否把字符串拆分为字典里的单词 @LeetCode 只不过要求计算的并不不过能否拆分，而是要求出全部的拆分方案。

因此用递归。

可是直接递归做会超时，原因是LeetCode里有几个非常长可是无法拆分的情况。所以就先跑一遍Word Break，先推断能否拆分。然后再进行拆分。

递归思路就是，逐一尝试字典里的每个单词，看看哪一个单词和S的开头部分匹配，假设匹配则递归处理S的除了开头部分，直到S为空。说明能够匹配。

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"].

public class Solution {
    public List<String> wordBreak(String s, Set<String> dict) {
        List<String> list = new ArrayList<String>();
        List<String> ret = new ArrayList<String>();
        rec(s, dict, list, ret);
        return ret;
    }
    
    public void rec(String s, Set<String> dict, List<String> list, List<String> ret) {
        if(!isBreak(s, dict)){  // test before run to avoid TLE
            return;
        }
        if(s.length() == 0) {
            String concat = "";
            for(int i=0; i<list.size(); i++) {
                concat += list.get(i);
                if(i != list.size()-1) {
                    concat += " ";
                }
            }
            ret.add(concat);
            return;
        }
        
        for(String cur : dict) {
        	if(cur.length() > s.length()) {     // avoid out of boundary
        		continue;
        	}
            String substr = s.substring(0, cur.length());
            if(substr.equals(cur)) {
                list.add(substr);
                rec(s.substring(cur.length()), dict, list, ret);
                list.remove(list.size()-1);
            }
        }
    }
    
    public boolean isBreak(String s, Set<String> dict) {
        boolean[] canBreak = new boolean[s.length()+1];
        canBreak[0] = true;
        
        for(int i=1; i<=s.length(); i++) {
            boolean flag = false;
            for(int j=0; j<i; j++) {
                if(canBreak[j] && dict.contains(s.substring(j,i))) {
                    flag = true;
                    break;
                }
            }
            canBreak[i] = flag;
        }
        return canBreak[s.length()];
    }
}