LeetCode 140. Word Break II

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

Note:

The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.

Example 1:

Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]

Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.
 

Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

不解釋，程式碼如下：

    public List<String> wordBreak(String s, List<String> wordDict) {
        HashSet<String> set = new HashSet<>();
        for (String ss : wordDict) {
            set.add(ss);
        }
        boolean[] breakable = new boolean[s.length() + 1];
        breakable[0] = true;
        for (int i = 0; i < s.length(); i++) {
            for (int p = 0; p <=i; p++) {
                String curS = s.substring(p, i+1);
                if (set.contains(curS) && breakable[p]) {
                    breakable[i+1] = true;
                }
            }
        }
        if (!breakable[s.length()]) {
            return new ArrayList<>();
        }
        List<List<String>> allResult = new ArrayList<>();
        for (int i = 0; i < s.length(); i++) {
            List<String> res = new ArrayList<>();
            for (int p = 0; p <= i; p++) {
                String curs = s.substring(p, i + 1);
                if (set.contains(curs)) {
                    if (p == 0) {
                        res.add(curs);
                    } else if (allResult.get(p-1).size() != 0) {
                        for (String ss : allResult.get(p-1)) {
                            ss = ss + " " + s.substring(p, i + 1);
                            res.add(ss);
                        }
                    } else {
                        continue;
                    }
                }
            }
            allResult.add(res);
        }
        return allResult.get(allResult.size() - 1);
    }