diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/Subsets.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/T08_Subsets.java
similarity index 98%
rename from Leecode/src/main/java/com/markilue/leecode/backtrace/Subsets.java
rename to Leecode/src/main/java/com/markilue/leecode/backtrace/T08_Subsets.java
index 77367b1..c6e92ee 100644
--- a/Leecode/src/main/java/com/markilue/leecode/backtrace/Subsets.java
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/T08_Subsets.java
@@ -16,7 +16,7 @@ import java.util.List;
  *  解集 不能 包含重复的子集。你可以按 任意顺序 返回解集。
  * @Version: 1.0
  */
-public class Subsets {
+public class T08_Subsets {
 
     @Test
     public void test(){
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/SubsetsWithDup.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/T09_SubsetsWithDup.java
similarity index 99%
rename from Leecode/src/main/java/com/markilue/leecode/backtrace/SubsetsWithDup.java
rename to Leecode/src/main/java/com/markilue/leecode/backtrace/T09_SubsetsWithDup.java
index bea6c89..7d1768d 100644
--- a/Leecode/src/main/java/com/markilue/leecode/backtrace/SubsetsWithDup.java
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/T09_SubsetsWithDup.java
@@ -14,7 +14,7 @@ import java.util.*;
  * 解集 不能 包含重复的子集。返回的解集中，子集可以按 任意顺序 排列。
  * @Version: 1.0
  */
-public class SubsetsWithDup {
+public class T09_SubsetsWithDup {
 
     @Test
     public void test() {
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/FindSubsequences.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/T10_FindSubsequences.java
similarity index 99%
rename from Leecode/src/main/java/com/markilue/leecode/backtrace/FindSubsequences.java
rename to Leecode/src/main/java/com/markilue/leecode/backtrace/T10_FindSubsequences.java
index 8c5f89e..19c0602 100644
--- a/Leecode/src/main/java/com/markilue/leecode/backtrace/FindSubsequences.java
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/T10_FindSubsequences.java
@@ -17,7 +17,7 @@ import java.util.Set;
  * 数组中可能含有重复元素，如出现两个整数相等，也可以视作递增序列的一种特殊情况。
  * @Version: 1.0
  */
-public class FindSubsequences {
+public class T10_FindSubsequences {
 
 
     @Test
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/Permute.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/T11_Permute.java
similarity index 98%
rename from Leecode/src/main/java/com/markilue/leecode/backtrace/Permute.java
rename to Leecode/src/main/java/com/markilue/leecode/backtrace/T11_Permute.java
index e8dfcca..e3a330b 100644
--- a/Leecode/src/main/java/com/markilue/leecode/backtrace/Permute.java
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/T11_Permute.java
@@ -15,7 +15,7 @@ import java.util.List;
  *  给定一个不含重复数字的数组 nums ，返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。
  * @Version: 1.0
  */
-public class Permute {
+public class T11_Permute {
 
     @Test
     public void test(){
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/PermuteUnique.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/T12_PermuteUnique.java
similarity index 98%
rename from Leecode/src/main/java/com/markilue/leecode/backtrace/PermuteUnique.java
rename to Leecode/src/main/java/com/markilue/leecode/backtrace/T12_PermuteUnique.java
index ae7369a..b4242be 100644
--- a/Leecode/src/main/java/com/markilue/leecode/backtrace/PermuteUnique.java
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/T12_PermuteUnique.java
@@ -16,7 +16,7 @@ import java.util.List;
  * 给定一个可包含重复数字的序列 nums ，按任意顺序 返回所有不重复的全排列。
  * @Version: 1.0
  */
-public class PermuteUnique {
+public class T12_PermuteUnique {
 
     @Test
     public void test() {
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T08_Subsets.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T08_Subsets.java
new file mode 100644
index 0000000..daed256
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T08_Subsets.java
@@ -0,0 +1,52 @@
+package com.markilue.leecode.backtrace.second;
+
+import org.junit.Test;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.backtrace.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-02-02  10:07
+ *@Description:
+ * TODO  二刷力扣78题  子集:
+ *  给你一个整数数组 nums ，数组中的元素 互不相同 。返回该数组所有可能的子集（幂集）。
+ *  解集 不能 包含重复的子集。你可以按 任意顺序 返回解集。
+ *@Version: 1.0
+ */
+public class T08_Subsets {
+
+    @Test
+    public void test() {
+        int[] nums={1,2,3};
+        System.out.println(subsets(nums));
+    }
+
+    List<List<Integer>> result = new ArrayList<>();
+    List<Integer> cur = new ArrayList<>();
+
+    public List<List<Integer>> subsets(int[] nums) {
+        backtracking(nums, 0);
+        return result;
+
+    }
+
+    public void backtracking(int[] nums, int level) {
+        if (level == nums.length) {
+            result.add(new ArrayList<>(cur));
+            return;
+        }
+
+        for (int i = 0; i <= 1; i++) {
+            if (i == 0) {
+                backtracking(nums, level + 1);
+            } else {
+                cur.add(nums[level]);
+                backtracking(nums, level+1);
+                cur.remove(cur.size() - 1);
+            }
+        }
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T09_SubsetsWithDup.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T09_SubsetsWithDup.java
new file mode 100644
index 0000000..3b93f72
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T09_SubsetsWithDup.java
@@ -0,0 +1,100 @@
+package com.markilue.leecode.backtrace.second;
+
+import org.junit.Test;
+
+import java.util.*;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.backtrace.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-02-02  10:32
+ *@Description:
+ * TODO  力扣90题  子集II:
+ *  给你一个整数数组 nums ，其中可能包含重复元素，请你返回该数组所有可能的子集（幂集）。
+ *  解集 不能 包含重复的子集。返回的解集中，子集可以按 任意顺序 排列。
+ *@Version: 1.0
+ */
+public class T09_SubsetsWithDup {
+    @Test
+    public void test() {
+        int[] nums = {1, 2, 2};
+        System.out.println(subsetsWithDup1(nums));
+    }
+
+    /**
+     * 两种思路:1.树层去重2.map记录元素个数，横向遍历使用
+     * @param nums
+     * @return
+     */
+    Map<Integer, Integer> map = new HashMap<Integer, Integer>();//<num,count>
+    List<Integer> key = new ArrayList<>();
+    List<Integer> cur = new ArrayList<>();
+    List<List<Integer>> result = new ArrayList<>();
+
+    public List<List<Integer>> subsetsWithDup(int[] nums) {
+        for (int num : nums) {
+            Integer count = map.getOrDefault(num, 0);
+            if (count == 0) {
+                key.add(num);
+            }
+            map.put(num, count + 1);
+        }
+        backtracking(0);
+        return result;
+    }
+
+    /**
+     * map记录次数法
+     * @param level
+     */
+    public void backtracking(int level) {
+        if (level == key.size()) {
+            result.add(new ArrayList<>(cur));
+            return;
+        }
+        Integer keyNow = key.get(level);
+        Integer count = map.get(keyNow);
+        for (int i = 0; i <= count; i++) {
+            if (i != 0) {
+                cur.add(keyNow);
+            }
+            backtracking(level + 1);
+        }
+        for (int i = 0; i < count; i++) {
+            cur.remove(cur.size() - 1);
+        }
+    }
+
+
+    public List<List<Integer>> subsetsWithDup1(int[] nums) {
+        Arrays.sort(nums);
+        backtracking1(nums,0);
+        return result;
+    }
+
+    /**
+     * 树层去重法
+     * @param nums
+     */
+    public void backtracking1(int[] nums, int startIndex) {
+        result.add(new ArrayList<>(cur));
+        if (startIndex > nums.length) {
+            return;
+        }
+
+        for (int i = startIndex; i < nums.length; i++) {
+            //树层去重
+            if (i != startIndex && nums[i - 1] == nums[i]) {
+                continue;
+            }
+            cur.add(nums[i]);
+            backtracking1(nums, i + 1);
+            cur.remove(cur.size() - 1);
+        }
+
+
+    }
+
+
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T10_FindSubsequences.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T10_FindSubsequences.java
new file mode 100644
index 0000000..546180e
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T10_FindSubsequences.java
@@ -0,0 +1,101 @@
+package com.markilue.leecode.backtrace.second;
+
+import org.junit.Test;
+
+import java.util.ArrayList;
+import java.util.HashSet;
+import java.util.List;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.backtrace.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-02-02  11:28
+ *@Description:
+ * TODO  二刷力扣491  递增子序列:
+ *  给你一个整数数组 nums ，找出并返回所有该数组中不同的递增子序列，递增子序列中 至少有两个元素 。你可以按 任意顺序 返回答案。
+ *  数组中可能含有重复元素，如出现两个整数相等，也可以视作递增序列的一种特殊情况。
+ *@Version: 1.0
+ */
+public class T10_FindSubsequences {
+
+    @Test
+    public void test() {
+        int[] nums = {4, 6, 7, 7};
+        System.out.println(findSubsequences(nums));
+    }
+
+
+    List<Integer> cur = new ArrayList<>();
+    List<List<Integer>> result = new ArrayList<>();
+
+    /**
+     * 与上一题子集II类似，都是需要树层去重即可
+     * @param nums
+     * @return
+     */
+    public List<List<Integer>> findSubsequences(int[] nums) {
+        backtracking(nums, 0);
+        return result;
+
+    }
+
+    public void backtracking(int[] nums, int startIndex) {
+        if (cur.size() >= 2) {
+            result.add(new ArrayList<>(cur));
+        }
+        if (startIndex == nums.length) return;
+        HashSet<Integer> set = new HashSet<>();
+        for (int i = startIndex; i < nums.length; i++) {
+            //同一树层去除,由于相同数不一定连续,所以使用set来进行树层去重
+            //这里还可以使用数组来去重，当数组的指定数位置的值为1时则证明使用过
+            if (set.contains(nums[i])) {
+                continue;
+            }
+            set.add(nums[i]);
+            if (cur.isEmpty() || cur.get(cur.size() - 1) <= nums[i]) {
+                //是递增的
+                cur.add(nums[i]);
+                backtracking(nums, i + 1);
+                cur.remove(cur.size() - 1);
+            }
+        }
+
+    }
+
+
+    /**
+     * 官方最快  2ms,可能快在用lastNum来记录cur(last)
+     * @param nums
+     * @return
+     */
+    public List<List<Integer>> findSubsequences1(int[] nums) {
+        List<List<Integer>> list = new ArrayList<>();
+        if (nums == null || nums.length == 0) return list;
+        List<Integer> result = new ArrayList<>();
+
+        dfs(0, nums, result, list, -101);
+        return list;
+    }
+
+    private void dfs(int idx, int[] nums, List<Integer> result, List<List<Integer>> list, int lastNum) {
+        if (result.size() > 1) {
+            list.add(new ArrayList<>(result));
+        }
+
+        for (int i = idx; i < nums.length; i++) {
+            if (isRepeat(nums, idx, i)) continue;
+            if (nums[i] < lastNum) continue;
+            result.add(nums[i]);
+            dfs(i + 1, nums, result, list, nums[i]);
+            result.remove(result.size() - 1);
+        }
+    }
+
+    private boolean isRepeat(int[] nums, int idx, int i) {
+        for (int j = idx; j < i; j++) {
+            if (nums[j] == nums[i]) return true;
+        }
+        return false;
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T11_Permute.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T11_Permute.java
new file mode 100644
index 0000000..689ae92
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T11_Permute.java
@@ -0,0 +1,83 @@
+package com.markilue.leecode.backtrace.second;
+
+import com.sun.org.apache.bcel.internal.generic.LUSHR;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.backtrace.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-02-02  11:55
+ *@Description:
+ * TODO  力扣46题  全排列:
+ *  给定一个不含重复数字的数组 nums ，返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。
+ *@Version: 1.0
+ */
+public class T11_Permute {
+
+    List<Integer> cur = new ArrayList<>();
+    List<List<Integer>> result = new ArrayList<>();
+    boolean[] used;
+
+    /**
+     * 全排列的题，实际上就是顺序也有关系，所以直接不用传startIndex,每次都从0开始就好
+     * 把used放外面速度击败82.16%，内存击败28.47%  1ms
+     * @param nums
+     * @return
+     */
+    public List<List<Integer>> permute(int[] nums) {
+        used = new boolean[nums.length];
+        backtracking(nums);
+        return result;
+
+    }
+
+    public void backtracking(int[] nums) {
+        if(cur.size()==nums.length){
+            result.add(new ArrayList<>(cur));
+            return;
+        }
+
+        for (int i = 0; i < nums.length; i++) {
+            if(!used[i]){
+                used[i]=true;
+                cur.add(nums[i]);
+                backtracking(nums);
+                cur.remove(cur.size()-1);
+                used[i]=false;
+            }
+        }
+
+    }
+
+
+    /**
+     * 把used放里面传参
+     * 速度击败100%，内存击败83.99%  0ms
+     * @param nums
+     * @return
+     */
+    public List<List<Integer>> permute1(int[] nums) {
+        backtracking(nums,new boolean[nums.length]);
+        return result;
+    }
+    public void backtracking(int[] nums, boolean[] used) {
+        if(cur.size()==nums.length){
+            result.add(new ArrayList<>(cur));
+            return;
+        }
+
+        for (int i = 0; i < nums.length; i++) {
+            if(!used[i]){
+                used[i]=true;
+                cur.add(nums[i]);
+                backtracking(nums,used);
+                cur.remove(cur.size()-1);
+                used[i]=false;
+            }
+        }
+
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T12_PermuteUnique.java b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T12_PermuteUnique.java
new file mode 100644
index 0000000..939d530
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/backtrace/second/T12_PermuteUnique.java
@@ -0,0 +1,87 @@
+package com.markilue.leecode.backtrace.second;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.backtrace.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-02-02  12:08
+ *@Description:
+ * TODO  力扣47题  全排列II:
+ * 给定一个可包含重复数字的序列 nums ，按任意顺序 返回所有不重复的全排列。
+ *@Version: 1.0
+ */
+public class T12_PermuteUnique {
+
+    List<Integer> cur = new ArrayList<>();
+    List<List<Integer>> result = new ArrayList<>();
+
+    /**
+     * 要做的事似乎还是树层去重
+     * @param nums
+     * @return
+     */
+    public List<List<Integer>> permuteUnique(int[] nums) {
+        backtracking(nums, new boolean[nums.length]);
+        return result;
+    }
+
+    public void backtracking(int[] nums, boolean[] used) {
+        if (nums.length == cur.size()) {
+            result.add(new ArrayList<>(cur));
+            return;
+        }
+        boolean[] curUsed = new boolean[21];//本树层是否用过
+        for (int i = 0; i < nums.length; i++) {
+            if (curUsed[nums[i] + 10]) continue;
+            if (!used[i]) {
+                curUsed[nums[i] + 10] = true;
+                used[i] = true;
+                cur.add(nums[i]);
+                backtracking(nums, used);
+                cur.remove(cur.size() - 1);
+                used[i] = false;
+            }
+        }
+
+    }
+
+
+    /**
+     * 可以进一步优化为一个used数组
+     * @param nums
+     * @param
+     */
+    public List<List<Integer>> permuteUnique1(int[] nums) {
+        Arrays.sort(nums);
+        backtracking(nums, new boolean[nums.length]);
+        return result;
+    }
+    public void backtracking1(int[] nums, boolean[] used) {
+        if (nums.length == cur.size()) {
+            result.add(new ArrayList<>(cur));
+            return;
+        }
+
+        for (int i = 0; i < nums.length; i++) {
+            // used[i - 1] == true，说明同一树枝nums[i - 1]使用过
+            // used[i - 1] == false，说明同一树层nums[i - 1]使用过
+            // 如果同一树层nums[i - 1]使用过则直接跳过
+            if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
+                continue;
+            }
+            if (!used[i]) {
+                used[i] = true;
+                cur.add(nums[i]);
+                backtracking1(nums, used);
+                cur.remove(cur.size() - 1);
+                used[i] = false;
+            }
+        }
+
+    }
+}
+}