diff --git a/code/lc242.java b/code/lc242.java
new file mode 100644
index 0000000000000000000000000000000000000000..e75bbcca4d9e6ae99f62b36557548cdc454490d0
--- /dev/null
+++ b/code/lc242.java
@@ -0,0 +1,23 @@
+package code;
+/*
+ * 242. Valid Anagram
+ * 题意：字符串t是否为s打乱后的重排列
+ * 难度：Easy
+ * 分类：Hash Table, Sort
+ * 思路：
+ * Tips：
+ */
+public class lc242 {
+    public boolean isAnagram(String s, String t) {
+        if(s.length()!=t.length()) return false;
+        int[] chs = new int[26];
+        for(int i=0; i<s.length(); i++){
+            chs[s.charAt(i)-'a']++;
+        }
+        for(int i=0; i<t.length(); i++){
+            chs[t.charAt(i)-'a']--;
+            if(chs[t.charAt(i)-'a']<0) return false;
+        }
+        return true;
+    }
+}
diff --git a/code/lc268.java b/code/lc268.java
new file mode 100644
index 0000000000000000000000000000000000000000..abd255244811eb9043da48cf0aaf34fbd1fa037a
--- /dev/null
+++ b/code/lc268.java
@@ -0,0 +1,23 @@
+package code;
+/*
+ * 268. Missing Number
+ * 题意：找出 0~n 中少的那个数
+ * 难度：Easy
+ * 分类：Array, Math, Bit Manipulation
+ * 思路：两种巧妙的方法，时间空间都是O(1)
+ *      异或
+ *      求和以后，减去所有
+ * Tips：
+ */
+public class lc268 {
+    public int missingNumber(int[] nums) {
+        int res = nums.length*(nums.length+1)/2;
+        for(int i:nums) res-=i;
+        return res;
+    }
+    public int missingNumber2(int[] nums) {
+        int res = nums.length;
+        for(int i=0; i<nums.length; i++) res^=i^nums[i];
+        return res;
+    }
+}
\ No newline at end of file
diff --git a/code/lc326.java b/code/lc326.java
new file mode 100644
index 0000000000000000000000000000000000000000..6ffb51759acb027a79a0dd391b42e26e0666fedb
--- /dev/null
+++ b/code/lc326.java
@@ -0,0 +1,25 @@
+package code;
+/*
+ * 326. Power of Three
+ * 题意：判断该数是否为3的幂
+ * 难度：Easy
+ * 分类：Math
+ * 思路：除以3，除到不能整除，判断是否为1
+ * Tips：
+ */
+public class lc326 {
+    public boolean isPowerOfThree(int n) {
+        if (n < 1) return false;
+        while (n % 3 == 0) n /= 3;
+        return n == 1;
+    }
+    public boolean isPowerOfThree2(int n) {
+        if(n==1) return true;
+        double d = n;
+        while(d>1){
+            d=d/3;
+            if(d==1) return true;
+        }
+        return false;
+    }
+}
diff --git a/code/lc329.java b/code/lc329.java
new file mode 100644
index 0000000000000000000000000000000000000000..32382e31d73ebd1627edac68efdb28c7ae9d2999
--- /dev/null
+++ b/code/lc329.java
@@ -0,0 +1,33 @@
+package code;
+/*
+ * 329. Longest Increasing Path in a Matrix
+ * 题意：寻找最长的递增路径
+ * 难度：Hard
+ * 分类：Depth-first Search, Topological Sort, Memoization
+ * 思路：带记忆的dfs，之前计算的最优解可以直接合并
+ * Tips：
+ */
+public class lc329 {
+    public int longestIncreasingPath(int[][] matrix) {
+        if(matrix.length==0) return 0;
+        int[][] cache = new int[matrix.length][matrix[0].length];   //存储计算过的结果
+        int res = 0;
+        for (int i = 0; i < matrix.length ; i++) {
+            for (int j = 0; j < matrix[0].length ; j++) {
+                res = Math.max(dfs(matrix, i, j, cache), res);
+            }
+        }
+        return res;
+    }
+
+    public int dfs(int[][] matrix, int i, int j, int[][] cache){
+        int max = 1;
+        if(cache[i][j]!=0) return cache[i][j];
+        if( i>0 && matrix[i-1][j]>matrix[i][j]) max = Math.max(dfs(matrix, i-1, j, cache)+1, max);
+        if( j>0 && matrix[i][j-1]>matrix[i][j]) max = Math.max(dfs(matrix, i, j-1, cache)+1, max);
+        if( i+1<matrix.length && matrix[i+1][j]>matrix[i][j]) max = Math.max(dfs(matrix, i+1, j, cache)+1, max);
+        if( j+1<matrix[0].length && matrix[i][j+1]>matrix[i][j]) max = Math.max(dfs(matrix, i, j+1, cache)+1, max);
+        cache[i][j] = max;
+        return max;
+    }
+}