Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.
Note that the same word in the dictionary may be reused multiple times in the segmentation.
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: []
Constraints:
- 1 <= s.length <= 20
- 1 <= wordDict.length <= 1000
- 1 <= wordDict[i].length <= 10
- s and wordDict[i] consist of only lowercase English letters.
- All the strings of wordDict are unique.
class Solution {
public List<String> wordBreak(String s, List<String> wordDict) {
return DFS(s, wordDict, new HashMap<String, LinkedList<String>>());
}
// DFS function returns an array including all substrings derived from s.
List<String> DFS(String s, List<String> wordDict, HashMap<String, LinkedList<String>>map) {
if (map.containsKey(s))
return map.get(s);
LinkedList<String>res = new LinkedList<String>();
if (s.length() == 0) {
res.add("");
return res;
}
for (String word : wordDict) {
if (s.startsWith(word)) {
List<String>sublist = DFS(s.substring(word.length()), wordDict, map);
for (String sub : sublist)
res.add(word + (sub.isEmpty() ? "" : " ") + sub);
}
}
map.put(s, res);
return res;
}
}