Update 472._Concatenated_Words.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/472._Concatenated_Words.md

@@ -104,3 +104,36 @@ class Solution:
 ******- 时间复杂度: O(words_num * word_len)******- 空间复杂度: O(words_num * word_len)******
 
 
+将这道题转化为第139题的解法，一个word肯定只能由比它更短的word组成，因此我们先按照word长度将words排个序
+
+然后遍历，看当前的word是否能由它之前的words组成，这个地方就是第139题的解法了，但是稍微有一点点区别，那就是我们这里如果当前word必须由两个及两个以上word组成才可以，也就是说如果当前word的长度小于2的话，肯定不能由两个及两个以上word组成
+
+beats 8.57%
+
+```python
+class Solution:
+    def findAllConcatenatedWordsInADict(self, words):
+        """
+        :type words: List[str]
+        :rtype: List[str]
+        """
+        if not words or len(words) == 0:
+            return []
+        res = []
+        words.sort(key = lambda s: len(s))
+        pre_words_lookup = set()
+        for i in range(len(words)):
+            if self.canConcatenated(words[i], pre_words_lookup):
+                res.append(words[i])
+            pre_words_lookup.add(words[i])
+        return res
+    
+    def canConcatenated(self, s, wordDict):
+        if len(s) <= 1:
+            return False
+        ok = [True]
+        for i in range(1, len(s)+1):
+            ok += [any(ok[j] and s[j:i] in wordDict for j in range(i))]
+        return ok[-1]
+```
+