diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md
index 99577f0c..417f4a6d 100644
--- a/problems/0040.组合总和II.md
+++ b/problems/0040.组合总和II.md
@@ -258,40 +258,45 @@ public:
 **使用标记数组**
 ```Java
 class Solution {
-    List<List<Integer>> lists = new ArrayList<>();
-    Deque<Integer> deque = new LinkedList<>();
-    int sum = 0;
+  LinkedList<Integer> path = new LinkedList<>();
+  List<List<Integer>> ans = new ArrayList<>();
+  boolean[] used;
+  int sum = 0;
 
-    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
-        //为了将重复的数字都放到一起，所以先进行排序
-        Arrays.sort(candidates);
-        //加标志数组，用来辅助判断同层节点是否已经遍历
-        boolean[] flag = new boolean[candidates.length];
-        backTracking(candidates, target, 0, flag);
-        return lists;
-    }
+  public List<List<Integer>> combinationSum2(int[] candidates, int target) {
+    used = new boolean[candidates.length];
+    // 加标志数组，用来辅助判断同层节点是否已经遍历
+    Arrays.fill(used, false);
+    // 为了将重复的数字都放到一起，所以先进行排序
+    Arrays.sort(candidates);
+    backTracking(candidates, target, 0);
+    return ans;
+  }
 
-    public void backTracking(int[] arr, int target, int index, boolean[] flag) {
-        if (sum == target) {
-            lists.add(new ArrayList(deque));
-            return;
-        }
-        for (int i = index; i < arr.length && arr[i] + sum <= target; i++) {
-            //出现重复节点，同层的第一个节点已经被访问过，所以直接跳过
-            if (i > 0 && arr[i] == arr[i - 1] && !flag[i - 1]) {
-                continue;
-            }
-            flag[i] = true;
-            sum += arr[i];
-            deque.push(arr[i]);
-            //每个节点仅能选择一次，所以从下一位开始
-            backTracking(arr, target, i + 1, flag);
-            int temp = deque.pop();
-            flag[i] = false;
-            sum -= temp;
-        }
+  private void backTracking(int[] candidates, int target, int startIndex) {
+    if (sum == target) {
+      ans.add(new ArrayList(path));
     }
+    for (int i = startIndex; i < candidates.length; i++) {
+      if (sum + candidates[i] > target) {
+        break;
+      }
+      // 出现重复节点，同层的第一个节点已经被访问过，所以直接跳过
+      if (i > 0 && candidates[i] == candidates[i - 1] && !used[i - 1]) {
+        continue;
+      }
+      used[i] = true;
+      sum += candidates[i];
+      path.add(candidates[i]);
+      // 每个节点仅能选择一次，所以从下一位开始
+      backTracking(candidates, target, i + 1);
+      used[i] = false;
+      sum -= candidates[i];
+      path.removeLast();
+    }
+  }
 }
+
 ```
 **不使用标记数组**
 ```Java
diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md
index 1ef278ff..964facee 100644
--- a/problems/0216.组合总和III.md
+++ b/problems/0216.组合总和III.md
@@ -260,6 +260,37 @@ class Solution {
 		}
 	}
 }
+
+// 上面剪枝 i <= 9 - (k - path.size()) + 1; 如果还是不清楚
+// 也可以改为 if (path.size() > k) return; 执行效率上是一样的
+class Solution {
+    LinkedList<Integer> path = new LinkedList<>();
+    List<List<Integer>> ans = new ArrayList<>();
+    public List<List<Integer>> combinationSum3(int k, int n) {
+        build(k, n, 1, 0);
+        return ans;
+    }
+
+    private void build(int k, int n, int startIndex, int sum) {
+        
+        if (sum > n) return;
+
+        if (path.size() > k) return;
+
+        if (sum == n && path.size() == k) {
+            ans.add(new ArrayList<>(path));
+            return;
+        }
+        
+        for(int i = startIndex; i <= 9; i++) {
+            path.add(i);
+            sum += i;
+            build(k, n, i + 1, sum);
+            sum -= i;
+            path.removeLast();
+        }
+    }
+}
 ```
 
 其他方法