[LeetCode] 140. Word Break II 单词拆分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:
[]

139. Word Break 的拓展，那道题只是让判断是否可以拆分，而这道题要求输出所有可能的拆分组合。

解法：dp + dfs，先用dp计算是否可以拆分，然后用dfs求出具体拆分的组合。

Java:

public static List<String> wordBreak(String s, Set<String> dict) {
    //create an array of ArrayList<String>
    List<String> dp[] = new ArrayList[s.length()+1];
    dp[0] = new ArrayList<String>();
 
    for(int i=0; i<s.length(); i++){
        if( dp[i] == null ) 
            continue; 
 
        for(String word:dict){
            int len = word.length();
            int end = i+len;
            if(end > s.length()) 
                continue;
 
            if(s.substring(i,end).equals(word)){
                if(dp[end] == null){
                    dp[end] = new ArrayList<String>();
                }
                dp[end].add(word);
            }
        }
    }
 
    List<String> result = new LinkedList<String>();
    if(dp[s.length()] == null) 
        return result; 
 
    ArrayList<String> temp = new ArrayList<String>();
    dfs(dp, s.length(), result, temp);
 
    return result;
}
 
public static void dfs(List<String> dp[],int end,List<String> result, ArrayList<String> tmp){
    if(end <= 0){
        String path = tmp.get(tmp.size()-1);
        for(int i=tmp.size()-2; i>=0; i--){
            path += " " + tmp.get(i) ;
        }
 
        result.add(path);
        return;
    }
 
    for(String str : dp[end]){
        tmp.add(str);
        dfs(dp, end-str.length(), result, tmp);
        tmp.remove(tmp.size()-1);
    }
}

Java:

public List<String> wordBreak(String s, Set<String> wordDict) {
    ArrayList<String> [] pos = new ArrayList[s.length()+1];
    pos[0]=new ArrayList<String>();
 
    for(int i=0; i<s.length(); i++){
        if(pos[i]!=null){
            for(int j=i+1; j<=s.length(); j++){
                String sub = s.substring(i,j);
                if(wordDict.contains(sub)){
                    if(pos[j]==null){
                        ArrayList<String> list = new ArrayList<String>();
                        list.add(sub);
                        pos[j]=list;
                    }else{
                        pos[j].add(sub);
                    }
 
                }
            }
        }
    }
 
    if(pos[s.length()]==null){
        return new ArrayList<String>();
    }else{
        ArrayList<String> result = new ArrayList<String>();
        dfs(pos, result, "", s.length());
        return result;
    }
}
 
public void dfs(ArrayList<String> [] pos, ArrayList<String> result, String curr, int i){
    if(i==0){
        result.add(curr.trim());
        return;
    }
 
    for(String s: pos[i]){
        String combined = s + " "+ curr;
        dfs(pos, result, combined, i-s.length());
    }
}　

Java:

public class Solution {
    Map<String, List<String>> done = new HashMap<>();
    Set<String> dict;

    public List<String> wordBreak(String s, Set<String> dict) {
        this.dict = dict;
        done.put("", new ArrayList<>());
        done.get("").add("");

        return dfs(s);
    }

    List<String> dfs(String s) {
        if (done.containsKey(s)) {
            return done.get(s);
        }
        List<String> ans = new ArrayList<>();

        for (int len = 1; len <= s.length(); len++) { 
            String s1 = s.substring(0, len);
            String s2 = s.substring(len);

            if (dict.contains(s1)) {
                List<String> s2_res = dfs(s2);
                for (String item : s2_res) {
                    if (item == "") {
                        ans.add(s1);
                    } else {
                        ans.add(s1 + " " + item);
                    }
                }
            }
        }
        done.put(s, ans);
        return ans;
    }
}　

Python:

class Solution(object):
    def wordBreak(self, s, wordDict):
        """
        :type s: str
        :type wordDict: Set[str]
        :rtype: List[str]
        """
        n = len(s)

        max_len = 0
        for string in wordDict:
            max_len = max(max_len, len(string))

        can_break = [False for _ in xrange(n + 1)]
        valid = [[False] * n for _ in xrange(n)]
        can_break[0] = True
        for i in xrange(1, n + 1):
            for l in xrange(1, min(i, max_len) + 1):
                if can_break[i-l] and s[i-l:i] in wordDict:
                    valid[i-l][i-1] = True
                    can_break[i] = True

        result = []
        if can_break[-1]:
            self.wordBreakHelper(s, valid, 0, [], result)
        return result

    def wordBreakHelper(self, s, valid, start, path, result):
        if start == len(s):
            result.append(" ".join(path))
            return
        for i in xrange(start, len(s)):
            if valid[start][i]:
                path += [s[start:i+1]]
                self.wordBreakHelper(s, valid, i + 1, path, result)
                path.pop()　

C++:

class Solution {
public:
    vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
        vector<string> res;
        string out;
        vector<bool> possible(s.size() + 1, true);
        wordBreakDFS(s, wordDict, 0, possible, out, res);
        return res;
    }
    void wordBreakDFS(string &s, unordered_set<string> &wordDict, int start, vector<bool> &possible, string &out, vector<string> &res) {
        if (start == s.size()) {
            res.push_back(out.substr(0, out.size() - 1));
            return;
        }
        for (int i = start; i < s.size(); ++i) {
            string word = s.substr(start, i - start + 1);
            if (wordDict.find(word) != wordDict.end() && possible[i + 1]) {
                out.append(word).append(" ");
                int oldSize = res.size();
                wordBreakDFS(s, wordDict, i + 1, possible, out, res);
                if (res.size() == oldSize) possible[i + 1] = false;
                out.resize(out.size() - word.size() - 1);
            }
        }
    }
};

　　

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[LeetCode] 139. Word Break 单词拆分

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