algorithm020
diff --git a/‎.idea/workspace.xml‎
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diff --git a/‎Week2/README.md‎
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diff --git a/‎Week2/__init__.py‎ b/‎Week2/__init__.py‎
diff --git a/‎Week2/easy/__init__.py‎ b/‎Week2/easy/__init__.py‎
diff --git a/‎Week2/easy/isAnagram.py‎
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@@ -6,4 +6,3 @@
     2) 树的节点 + (左子树) + (右子树), 左子树 = 左子树节点 + (subleftroot.left) + (subleftroot.right), 右子树 = 右子树节点 + (subrightroot.left) + (subrightroot.right),
        而所有节点都是同一属性的TreeNode, 这就是一个层层嵌套的结构，一个树是含有重复嵌套结构或者说重复子问题的结构。
     3）在实际的一些应用场景，解决树的一些问题例如插入、删除操作需要先遍历再做访问，在这里可以使用循环的方法，但是要借助栈做缓存保留遍历的记忆，这个本质上还是和递归里层层走进嵌入的模式是一样的。
-                     
@@ -0,0 +1,53 @@
+# ============================================
+# Leetco 242. 有效的字母异位词
+# 审题: 1. 进阶的情况: 如果输入字符串包含 unicode 字符怎么办？你能否调整你的解法来应对这种情况
+# not seen before
+# 解体思路: 字典形式储存词频
+# time complexity: O(2n)≈O(n)
+# ============================================
+
+
+# TODO Solution1
+#执行用时：48 ms, 在所有 Python 提交中击败了51.42%的用户
+#内存消耗：14.9 MB, 在所有 Python 提交中击败了11.08%的用户
+class Solution1(object):
+    def isAnagram(self, s, t):
+        """
+        :type s: str
+        :type t: str
+        :rtype: bool
+        """
+        # record the frequency/count of letters in s by key-value data structure
+        dict_ct = {}
+        if len(s) != len(t):
+            return False
+        for i in range(len(s)):
+            if s[i] not in dict_ct:
+                dict_ct[s[i]] = 0
+            dict_ct[s[i]] += 1
+        for i in range(len(t)):
+            if t[i] in dict_ct:
+                dict_ct[t[i]] -= 1
+            else:
+                return False
+
+            if dict_ct[t[i]] == 0:
+                del dict_ct[t[i]]
+
+        return (not dict_ct)
+
+
+# TODO 内存还是没有上来
+class Solution2(object):
+    def isAnagram(self, s, t):
+        """
+        :type s: str
+        :type t: str
+        :rtype: bool
+        """
+        if len(s) != len(s):
+            return False
+        if sorted(s) == sorted(t):
+            return True
+        else:
+            return False