140. Word Break II

╰半夏微凉° 2022-02-13 11:11 156阅读 0赞

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:
    []

与上一题不同，这次需要返回所有被分割的字符串。显然动态规划已经不适合这里了，这里我们可以用DFS的方法来解答。

public List<String> wordBreak(String s, List<String> wordDict) {
            return DFS(s, wordDict, new HashMap<>());
        }
    
        private List<String> DFS(String s, List<String> wordDict, Map<String, List<String>> map) {
            if(map.containsKey(s)) {//这里可以提供缓存效果
                return map.get(s);
            }
            List<String> ans = new LinkedList<>();
            if(s.length() == 0) {//s.length()=0表示已达到""可以被分割为""
                ans.add("");
                return ans;
            }
            for(int i = 0; i < wordDict.size(); ++i) {
                if(s.startsWith(wordDict.get(i))) {//对字典中每一个满足要求的字符串都查找s.subString(wordDict.get(i).length())的满足要求的字符串集
                    List<String> temp = DFS(s.substring(wordDict.get(i).length()), wordDict, map);
                    for(int j = 0; j < temp.size(); ++j) {
                        ans.add(wordDict.get(i) + (temp.get(j).isEmpty() ? "" : " ") + temp.get(j));
                    }
                }
            }
            map.put(s, ans);
            return ans;
        }