diff --git a/Leecode/src/main/java/com/markilue/leecode/dynamic/T13_CombinationSum4.java b/Leecode/src/main/java/com/markilue/leecode/dynamic/T13_CombinationSum4.java
index 9954306..e2070e1 100644
--- a/Leecode/src/main/java/com/markilue/leecode/dynamic/T13_CombinationSum4.java
+++ b/Leecode/src/main/java/com/markilue/leecode/dynamic/T13_CombinationSum4.java
@@ -2,6 +2,8 @@ package com.markilue.leecode.dynamic;
 
 import org.junit.Test;
 
+import java.util.Arrays;
+
 /**
  * @BelongsProject: Leecode
  * @BelongsPackage: com.markilue.leecode.dynamic
@@ -38,6 +40,7 @@ public class T13_CombinationSum4 {
      *              (3)dp初始化:dp[i][0]=1且dp[0][j%num[0]==0]=1
      *              (4)dp遍历顺序:由于是排列问题，因此先背包后num[i]
      *              (5)dp举例推导:
+     *       速度击败97.72%  内存击败90.4%
      * @param nums
      * @param target
      * @return
@@ -58,4 +61,33 @@ public class T13_CombinationSum4 {
         return dp[target];
 
     }
+
+
+    /**
+     * 官方最快:记忆化搜索:回溯+记忆化搜索
+     * 速度击败100%  内存击败96%
+     * @param nums
+     * @param target
+     * @return
+     */
+    public int combinationSum41(int[] nums, int target) {
+        int[] memo = new int[target + 1];
+        Arrays.fill(memo, -1);
+        memo[0] = 1;
+        return search(memo, nums, target);
+    }
+
+    public int search(int[] memo, int[] nums, int target){
+        if(memo[target] != -1){
+            return memo[target];
+        }
+        int res = 0;
+        for(int i : nums){
+            if(target >= i){
+                res += search(memo, nums, target - i);
+            }
+        }
+        memo[target] = res;//全部遍历完就填充记忆；避免下一次继续计算
+        return res;
+    }
 }
diff --git a/Leecode/src/main/java/com/markilue/leecode/dynamic/second/T13_CombinationSum4.java b/Leecode/src/main/java/com/markilue/leecode/dynamic/second/T13_CombinationSum4.java
new file mode 100644
index 0000000..180a372
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/dynamic/second/T13_CombinationSum4.java
@@ -0,0 +1,53 @@
+package com.markilue.leecode.dynamic.second;
+
+import org.junit.Test;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.dynamic.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-02-18  19:06
+ *@Description:
+ * TODO 力扣377题  组合总和Iv:
+ *  给你一个由 不同 整数组成的数组 nums ，和一个目标整数 target 。请你从 nums 中找出并返回总和为 target 的元素组合的个数。
+ *  题目数据保证答案符合 32 位整数范围。
+ *@Version: 1.0
+ */
+public class T13_CombinationSum4 {
+
+    @Test
+    public void test() {
+        int[] nums = {1, 2, 3};
+        int target = 4;
+        System.out.println(combinationSum4(nums, target));
+    }
+
+
+    /**
+     * 思路:还是完全背包问题:数字可以无限使用,但是不同的是:顺序不同算不同的组合，所以nums遍历需要放里面
+     *   TODO DP五部曲:
+     *      1.dp定义:  dp[i][j]表示使用nums[0-i]可能凑出j的个数
+     *      2.dp状态转移方程:  dp[i][j]  可以使用当前的数  或者  不使用当前的数
+     *                      dp[i][j]=dp[i][j-nums[i]]+dp[i-1][j]
+     *      3.dp初始化:  dp[0][j%nums[0]]=1
+     *      4.dp遍历顺序: 由于顺序不同算不同的组合:所以先从上到下；再从前往后
+     *      5.dp举例推导:
+     * @param nums
+     * @param target
+     * @return
+     */
+    public int combinationSum4(int[] nums, int target) {
+
+        int[] dp = new int[target + 1];
+        dp[0] = 1;
+
+        for (int j = 0; j < dp.length; j++) {
+            for (int i = 0; i < nums.length; i++) {
+                if (j >= nums[i]) dp[j] += dp[j - nums[i]];
+            }
+        }
+        return dp[target];
+
+
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/greedy/second/T02_WiggleMaxLength.java b/Leecode/src/main/java/com/markilue/leecode/greedy/second/T02_WiggleMaxLength.java
index 5174daa..68c0046 100644
--- a/Leecode/src/main/java/com/markilue/leecode/greedy/second/T02_WiggleMaxLength.java
+++ b/Leecode/src/main/java/com/markilue/leecode/greedy/second/T02_WiggleMaxLength.java
@@ -79,7 +79,7 @@ public class T02_WiggleMaxLength {
         int curdiff = 0;  //记录现在的差值
         int prediff = 0;  //记录上次的差值
         for (int i = 0; i < nums.length - 1; i++) {
-            curdiff = nums[i + 1] - nums[i];
+            curdiff = nums[i + 1] - nums[i] ;
             if ((prediff >= 0 && curdiff < 0) || (prediff <= 0 && curdiff > 0)) {
                 result++;
                 prediff = curdiff;