Word Break II 求把字符串拆分为字典里的单词的所有方案 @LeetCode

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这道题类似 Word Break 判断是否能把字符串拆分为字典里的单词 @LeetCode 只不过要求计算的并不仅仅是是否能拆分，而是要求出所有的拆分方案。因此用递归。

但是直接递归做会超时，原因是LeetCode里有几个很长但是无法拆分的情况，所以就先跑一遍Word Break，先判断能否拆分，然后再进行拆分。

递归思路就是，逐一尝试字典里的每一个单词，看看哪一个单词和S的开头部分匹配，如果匹配则递归处理S的除了开头部分，直到S为空，说明可以匹配。

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()];    }}

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