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"].

【题意】

相比Word Break，这次不仅要判断字符串能否被分解，还要把所有的分解情况给找出来。这样的话，我在Word Break用过的暴力法就可以很容易得改成这道题。

==================== 暴力法、穷尽法 ====================

【Java代码】

class Solution {
    List<String> ret;
    
    public List<String>  wordBreak(String s, Set<String> dict) {
        String[] all = dict.toArray(new String[0]);
        
        ret = new ArrayList<String>();
        
        //如果s中有字母没在dict出现过，那么结果肯定为false
        //这段代码的作用就是防止暴力法超时，有点投机的做法，我在Word Break解题报告中已讲过
        for (int i = 0; i < s.length(); i++) {
            boolean flag = false;
            for (int j = 0; j < all.length; j++) {
                if (all[j].indexOf(s.charAt(i)) > -1) {
                    flag = true;
                    break;
                }
            }
            if (!flag) {
                return ret;
            } 
        }
        
        nextWord(0, s, all, "");
        return ret;
    }
    
    void nextWord(int pos, String s, String[] all, String sent) {
        if (pos == s.length()) {
            ret.add(sent.trim());
        }
        
        for (int i = 0; i < all.length; i++) {
            if (s.indexOf(all[i], pos) == pos) {
                String str = sent;
                str += all[i] + " ";
                nextWord(pos + all[i].length(), s, all, str);
            }
        }
    }
}

【LeetCode】Word Break II 解题报告