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[leetcode] problem 140 – word break ii

admin 11月 13, 2020 0

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

No.1

Input:
s = “catsanddog”
wordDict = [“cat”, “cats”, “and”, “sand”, “dog”]

Output:
[
“cats and dog”,
“cat sand dog”
]

No.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.

No.3

Input:
s = “catsandog”
wordDict = [“cats”, “dog”, “sand”, “and”, “cat”]

Output:
[]

Code

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private Map<String, List<String>> map = new HashMap<>();

public List<String> (String s, List<String> wordDict) {
    Set<String> dict = new HashSet<>();

    for (String word : wordDict)
        dict.add(word);

    return dfs(s, dict);
}

private List<String> dfs(String s, Set<String> dict) {
    if (map.containsKey(s))
        return map.get(s);

    List<String> result = new ArrayList<>();

    for (int i = 1; i <= s.length(); i++) {
        String word = s.substring(0, i);

        if (!dict.contains(word))
            continue;

        if (i == s.length())
            result.add(s);
        else {
            List<String> segList = dfs(s.substring(i), dict);

            for (String seg : segList)
                result.add(word + " " + seg);
        }
    }

    map.put(s, result);
    return result;
}
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