diff --git a/src/main/java/com/code/leet/week/Solution260.java b/src/main/java/com/code/leet/week/Solution260.java
new file mode 100644
index 0000000..a605f31
--- /dev/null
+++ b/src/main/java/com/code/leet/week/Solution260.java
@@ -0,0 +1,94 @@
+package com.code.leet.week;
+
+
+import com.code.leet.entiy.TwoArray;
+
+/**
+ * @description:
+ * @author: Administrator
+ * @date: 2021/8/22-10:29
+ */
+public class Solution260 {
+    public static void main(String[] args) {
+        Solution260 solution = new Solution260();
+        TwoArray twoArray = new TwoArray("[[1,3,1,15],[1,3,3,1]]", true);
+        solution.gridGame(twoArray.getArr());
+    }
+
+    public int maximumDifference(int[] nums) {
+        int max = 0;
+        for (int i = 0; i < nums.length; i++) {
+            for (int j = i + 1; j < nums.length; j++) {
+                max = Math.max(max, nums[j] - nums[i]);
+            }
+        }
+        return max > 0 ? max : -1;
+    }
+
+    public long gridGame(int[][] grid) {
+        int size = grid[0].length;
+        long[][] sums = new long[2][size + 1];
+        for (int i = 0; i < size; i++) {
+            sums[0][i + 1] = sums[0][i] + grid[0][i];
+            sums[1][i + 1] = sums[1][i] + grid[1][i];
+        }
+        long min = Long.MAX_VALUE;
+        for (int i = 1; i < size + 1; i++) {
+            min = Math.min(min, Math.max(sums[0][size] - sums[0][i], sums[1][0] - sums[1][i - 1]));
+        }
+        return min;
+    }
+
+    public boolean placeWordInCrossword(char[][] board, String word) {
+        int size = word.length();
+        for (int i = 0; i < board.length; i++) {
+            int blank = 0;
+            boolean add = false;
+            for (int j = 0; j < board[0].length; j++) {
+                if (size - blank > board[0].length - j) {
+                    break;
+                }
+                if (board[i][j] == ' ' && add) {
+                    blank++;
+                } else if (board[i][j] == ' ' && !add && ((j > 1 && board[i][j - 1] == '#') || j == 0)) {
+                    if (blank == size) {
+                        return true;
+                    }
+                    add = true;
+                    blank = 0;
+                } else if (board[i][j] != ' ' && board[i][j] != '#') {
+                    if(word.contains(""+board[i][j])){
+                        String temp = word;
+                        int index = temp.indexOf(""+board[i][j]);
+                        if(blank==index){
+                            blank++;
+                        }
+                    }
+                } else {
+                    add = false;
+                }
+            }
+        }
+        for (int i = 0; i < board[0].length; i++) {
+            int blank = 0;
+            boolean add = false;
+            for (int j = 0; j < board.length; j++) {
+                if (size - blank > board.length - j) {
+                    break;
+                }
+                if (board[j][i] == ' ' && add) {
+                    blank++;
+                } else if (board[j][i] == ' ' && !add && ((i > 1 && board[j][i - 1] == '#') || i == 0)) {
+                    if (blank == size) {
+                        return true;
+                    }
+                    add = true;
+                    blank = 0;
+                } else {
+                    add = false;
+                }
+            }
+        }
+        return false;
+    }
+}
diff --git a/src/main/java/leetcode/editor/cn/DecodeWaysIi.java b/src/main/java/leetcode/editor/cn/DecodeWaysIi.java
new file mode 100644
index 0000000..716298e
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/DecodeWaysIi.java
@@ -0,0 +1,126 @@
+//一条包含字母 A-Z 的消息通过以下的方式进行了编码： 
+//
+// 
+//'A' -> 1
+//'B' -> 2
+//...
+//'Z' -> 26
+// 
+//
+// 要 解码 一条已编码的消息，所有的数字都必须分组，然后按原来的编码方案反向映射回字母（可能存在多种方式）。例如，"11106" 可以映射为： 
+//
+// 
+// "AAJF" 对应分组 (1 1 10 6) 
+// "KJF" 对应分组 (11 10 6) 
+// 
+//
+// 注意，像 (1 11 06) 这样的分组是无效的，因为 "06" 不可以映射为 'F' ，因为 "6" 与 "06" 不同。 
+//
+// 除了 上面描述的数字字母映射方案，编码消息中可能包含 '*' 字符，可以表示从 '1' 到 '9' 的任一数字（不包括 '0'）。例如，编码字符串 "1*
+//" 可以表示 "11"、"12"、"13"、"14"、"15"、"16"、"17"、"18" 或 "19" 中的任意一条消息。对 "1*" 进行解码，相当于解码
+//该字符串可以表示的任何编码消息。 
+//
+// 给你一个字符串 s ，由数字和 '*' 字符组成，返回 解码 该字符串的方法 数目 。 
+//
+// 由于答案数目可能非常大，返回对 10⁹ + 7 取余 的结果。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：s = "*"
+//输出：9
+//解释：这一条编码消息可以表示 "1"、"2"、"3"、"4"、"5"、"6"、"7"、"8" 或 "9" 中的任意一条。
+//可以分别解码成字符串 "A"、"B"、"C"、"D"、"E"、"F"、"G"、"H" 和 "I" 。
+//因此，"*" 总共有 9 种解码方法。
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：s = "1*"
+//输出：18
+//解释：这一条编码消息可以表示 "11"、"12"、"13"、"14"、"15"、"16"、"17"、"18" 或 "19" 中的任意一条。
+//每种消息都可以由 2 种方法解码（例如，"11" 可以解码成 "AA" 或 "K"）。
+//因此，"1*" 共有 9 * 2 = 18 种解码方法。
+// 
+//
+// 示例 3： 
+//
+// 
+//输入：s = "2*"
+//输出：15
+//解释：这一条编码消息可以表示 "21"、"22"、"23"、"24"、"25"、"26"、"27"、"28" 或 "29" 中的任意一条。
+//"21"、"22"、"23"、"24"、"25" 和 "26" 由 2 种解码方法，但 "27"、"28" 和 "29" 仅有 1 种解码方法。
+//因此，"2*" 共有 (6 * 2) + (3 * 1) = 12 + 3 = 15 种解码方法。
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 1 <= s.length <= 10⁵ 
+// s[i] 是 0 - 9 中的一位数字或字符 '*' 
+// 
+// Related Topics 字符串 动态规划 👍 137 👎 0
+
+package leetcode.editor.cn;
+
+//639:解码方法 II
+class DecodeWaysIi {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new DecodeWaysIi().new Solution();
+        solution.numDecodings("*1*1*0");
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        static final int MOD = 1000000007;
+
+        public int numDecodings(String s) {
+            int n = s.length();
+            long a = 0, b = 1, c = 0;
+            for (int i = 1; i <= n; ++i) {
+                c = b * check1digit(s.charAt(i - 1)) % MOD;
+                if (i > 1) {
+                    c = (c + a * check2digits(s.charAt(i - 2), s.charAt(i - 1))) % MOD;
+                }
+                a = b;
+                b = c;
+            }
+            return (int) c;
+        }
+
+        public int check1digit(char ch) {
+            if (ch == '0') {
+                return 0;
+            }
+            return ch == '*' ? 9 : 1;
+        }
+
+        public int check2digits(char c0, char c1) {
+            if (c0 == '*' && c1 == '*') {
+                return 15;
+            }
+            if (c0 == '*') {
+                return c1 <= '6' ? 2 : 1;
+            }
+            if (c1 == '*') {
+                if (c0 == '1') {
+                    return 9;
+                }
+                if (c0 == '2') {
+                    return 6;
+                }
+                return 0;
+            }
+            return (c0 != '0' && (c0 - '0') * 10 + (c1 - '0') <= 26) ? 1 : 0;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/PowerOfThree.java b/src/main/java/leetcode/editor/cn/PowerOfThree.java
new file mode 100644
index 0000000..3ca9ab6
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/PowerOfThree.java
@@ -0,0 +1,70 @@
+//给定一个整数，写一个函数来判断它是否是 3 的幂次方。如果是，返回 true ；否则，返回 false 。 
+//
+// 整数 n 是 3 的幂次方需满足：存在整数 x 使得 n == 3ˣ 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：n = 27
+//输出：true
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：n = 0
+//输出：false
+// 
+//
+// 示例 3： 
+//
+// 
+//输入：n = 9
+//输出：true
+// 
+//
+// 示例 4： 
+//
+// 
+//输入：n = 45
+//输出：false
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// -2³¹ <= n <= 2³¹ - 1 
+// 
+//
+// 
+//
+// 进阶： 
+//
+// 
+// 你能不使用循环或者递归来完成本题吗？ 
+// 
+// Related Topics 递归 数学 👍 184 👎 0
+
+package leetcode.editor.cn;
+
+//326:3的幂
+class PowerOfThree {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new PowerOfThree().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public boolean isPowerOfThree(int n) {
+            return n > 0 && 1162261467 % n == 0;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/doc/all.json b/src/main/java/leetcode/editor/cn/doc/all.json
index f27a358..ab494ae 100644
--- a/src/main/java/leetcode/editor/cn/doc/all.json
+++ b/src/main/java/leetcode/editor/cn/doc/all.json
@@ -1 +1,2 @@
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diff --git a/src/main/java/leetcode/editor/cn/doc/content/DecodeWaysIi.md b/src/main/java/leetcode/editor/cn/doc/content/DecodeWaysIi.md
new file mode 100644
index 0000000..8ee5e58
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/DecodeWaysIi.md
@@ -0,0 +1,65 @@
+<p>一条包含字母&nbsp;<code>A-Z</code> 的消息通过以下的方式进行了编码：</p>
+
+<pre>
+'A' -&gt; 1
+'B' -&gt; 2
+...
+'Z' -&gt; 26
+</pre>
+
+<p>要 <strong>解码</strong> 一条已编码的消息，所有的数字都必须分组，然后按原来的编码方案反向映射回字母（可能存在多种方式）。例如，<code>"11106"</code> 可以映射为：</p>
+
+<ul>
+	<li><code>"AAJF"</code> 对应分组 <code>(1 1 10 6)</code></li>
+	<li><code>"KJF"</code> 对应分组 <code>(11 10 6)</code></li>
+</ul>
+
+<p>注意，像 <code>(1 11 06)</code> 这样的分组是无效的，因为 <code>"06"</code> 不可以映射为 <code>'F'</code> ，因为 <code>"6"</code> 与 <code>"06"</code> 不同。</p>
+
+<p><strong>除了</strong> 上面描述的数字字母映射方案，编码消息中可能包含 <code>'*'</code> 字符，可以表示从 <code>'1'</code> 到 <code>'9'</code> 的任一数字（不包括 <code>'0'</code>）。例如，编码字符串 <code>"1*"</code> 可以表示 <code>"11"</code>、<code>"12"</code>、<code>"13"</code>、<code>"14"</code>、<code>"15"</code>、<code>"16"</code>、<code>"17"</code>、<code>"18"</code> 或 <code>"19"</code> 中的任意一条消息。对 <code>"1*"</code> 进行解码，相当于解码该字符串可以表示的任何编码消息。</p>
+
+<p>给你一个字符串 <code>s</code> ，由数字和 <code>'*'</code> 字符组成，返回 <strong>解码</strong> 该字符串的方法 <strong>数目</strong> 。</p>
+
+<p>由于答案数目可能非常大，返回对 <code>10<sup>9</sup> + 7</code> <strong>取余</strong> 的结果。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "*"
+<strong>输出：</strong>9
+<strong>解释：</strong>这一条编码消息可以表示 "1"、"2"、"3"、"4"、"5"、"6"、"7"、"8" 或 "9" 中的任意一条。
+可以分别解码成字符串 "A"、"B"、"C"、"D"、"E"、"F"、"G"、"H" 和 "I" 。
+因此，"*" 总共有 9 种解码方法。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "1*"
+<strong>输出：</strong>18
+<strong>解释：</strong>这一条编码消息可以表示 "11"、"12"、"13"、"14"、"15"、"16"、"17"、"18" 或 "19" 中的任意一条。
+每种消息都可以由 2 种方法解码（例如，"11" 可以解码成 "AA" 或 "K"）。
+因此，"1*" 共有 9 * 2 = 18 种解码方法。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "2*"
+<strong>输出：</strong>15
+<strong>解释：</strong>这一条编码消息可以表示 "21"、"22"、"23"、"24"、"25"、"26"、"27"、"28" 或 "29" 中的任意一条。
+"21"、"22"、"23"、"24"、"25" 和 "26" 由 2 种解码方法，但 "27"、"28" 和 "29" 仅有 1 种解码方法。
+因此，"2*" 共有 (6 * 2) + (3 * 1) = 12 + 3 = 15 种解码方法。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s[i]</code> 是 <code>0 - 9</code> 中的一位数字或字符 <code>'*'</code></li>
+</ul>
+<div><div>Related Topics</div><div><li>字符串</li><li>动态规划</li></div></div><br><div><li>👍 137</li><li>👎 0</li></div>
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diff --git a/src/main/java/leetcode/editor/cn/doc/content/PowerOfThree.md b/src/main/java/leetcode/editor/cn/doc/content/PowerOfThree.md
new file mode 100644
index 0000000..4419697
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/PowerOfThree.md
@@ -0,0 +1,50 @@
+<p>给定一个整数，写一个函数来判断它是否是 3 的幂次方。如果是，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p>
+
+<p>整数 <code>n</code> 是 3 的幂次方需满足：存在整数 <code>x</code> 使得 <code>n == 3<sup>x</sup></code></p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 27
+<strong>输出：</strong>true
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 0
+<strong>输出：</strong>false
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 9
+<strong>输出：</strong>true
+</pre>
+
+<p><strong>示例 4：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 45
+<strong>输出：</strong>false
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>-2<sup>31</sup> <= n <= 2<sup>31</sup> - 1</code></li>
+</ul>
+
+<p> </p>
+
+<p><strong>进阶：</strong></p>
+
+<ul>
+	<li>你能不使用循环或者递归来完成本题吗？</li>
+</ul>
+<div><div>Related Topics</div><div><li>递归</li><li>数学</li></div></div><br><div><li>👍 184</li><li>👎 0</li></div>
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