139. Word Break I & II

Word Break I

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given

s = "leetcode",

dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        if(s == null || dict == null) return false;
        boolean[] canSeg = new boolean[s.length()+1];
        // this is used when determine a whole word can be segmented or not
        canSeg[0] = true; 
        for(int i = 1; i <= s.length(); i++)
            for(int k = 0; k < i; k++)
                // if the whole word cannot be found in the dictionary, we can determine whether
                // all divided parts cannot be found in the dictionary.
                if(canSeg[k] && dict.contains(s.substring(k, i))) canSeg[i] = true;
        return canSeg[s.length()];
    }
}

上一个版本在面试讲解思路的时候，其实是非常麻烦的，下面这种方法使用的canSeg的长度是和s一样的，讲解思路的时候更容易，但是code稍微多一些。

public boolean wordBreak(String s, Set<String> dict) {
      if(s == null || dict == null) return false;
      boolean[] canBreak = new boolean[s.length()];
      // initial case
      if (dict.contains(s.substring(0, 1))) {
          canBreak[0] = true;
      }
      for (int i = 1; i < s.length(); i++) {
          if (dict.contains(s.substring(0, i + 1))) {
              canBreak[i] = true;
          } else {
              for (int j = 1; j <= i; j++) {
                  String sub = s.substring(j, i + 1);
                  if (canBreak[j - 1] && dict.contains(sub)) {
                      canBreak[i] = true;
                      break;
                  }
              }
          }
      }
      return canBreak[canBreak.length - 1];
  }

Word Break II

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"].

分析：

class Solution {
    public List<String> wordBreak(String s, List<String> wordDict) {
        return dfs(s, new HashSet<String>(wordDict), new HashMap<>());
    }       

    // DFS function returns an array including all substrings derived from s.
    List<String> dfs(String s, Set<String> wordDict, Map<String, List<String>>map) {
        if (map.containsKey(s)) {
            return map.get(s);
        }

        List<String> res = new LinkedList<>();
        
        for (int i = 1; i <= s.length(); i++) {
            String word = s.substring(0, i);
            if (wordDict.contains(word)) {
                if (i == s.length()) {
                    res.add(word);
                } else {
                    List<String> sublist = dfs(s.substring(i), wordDict, map);
                    for (String sub : sublist) {
                        res.add(word + " " + sub);
                    }
                }
            }
        }       
        map.put(s, res);
        return res;
    }
}