diff --git a/solution/2600-2699/2685.Count the Number of Complete Components/README.md b/solution/2600-2699/2685.Count the Number of Complete Components/README.md
index 523f109cef196..03951f996f45b 100644
--- a/solution/2600-2699/2685.Count the Number of Complete Components/README.md	
+++ b/solution/2600-2699/2685.Count the Number of Complete Components/README.md	
@@ -234,4 +234,61 @@ func countCompleteComponents(n int, edges [][]int) (ans int) {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### 方法二：取巧做法
+
+要解决的问题：
+
+1. 如何保存每一个节点与其它点联通状态
+2. 如何判断多个点是否是一个联通图
+
+对于第一点：实际上就是保存了当前到每个点的联通点集合（包括自己），方便后续判等。
+第二点：有了第一点之后，如果是连通图中的点就有：
+
+1. 此点包含此联通图中所有的点（包括自己）
+2. 并且只包含此联通图中的点
+
+拿示例一举例：
+
+-   5 包含的联通点有且只有自己，所以是连通图
+-   0 包含 0、1、2，同理 1、2 点也是
+-   3 和 4 也是包含自己和彼此
+-   基于以上就有以下代码实现：
+
+<!-- tabs:start -->
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int countCompleteComponents(int n, vector<vector<int>>& edges) {
+        int ans = 0;
+        vector<set<int>> m(n + 1, set<int>());
+        for (int i = 0; i < n; i++) {
+            m[i].insert(i);
+        }
+        for (auto x : edges) {
+            m[x[0]].insert(x[1]);
+            m[x[1]].insert(x[0]);
+        }
+        map<set<int>, int> s;
+        for (int i = 0; i < n; i++) {
+            s[m[i]] ++;
+        }
+        for (auto &[x,y] : s) {
+            if (y == x.size()) {
+                ans ++;
+            }
+        }
+        return ans;
+    }
+};
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->
diff --git a/solution/2600-2699/2685.Count the Number of Complete Components/README_EN.md b/solution/2600-2699/2685.Count the Number of Complete Components/README_EN.md
index fb6077f895077..f0292e5585d04 100644
--- a/solution/2600-2699/2685.Count the Number of Complete Components/README_EN.md	
+++ b/solution/2600-2699/2685.Count the Number of Complete Components/README_EN.md	
@@ -222,4 +222,61 @@ func countCompleteComponents(n int, edges [][]int) (ans int) {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 2: Simple Method
+
+Problems needed to solve：
+
+1. How do we maintain the link state between each node and the others? 如
+2. How can one determine whether multiple points form a connected graph?
+
+For the first one: we can maintain each node's connection set(including itself).
+
+For the second one: After solving the first one, we can see:
+
+-   the node itself includes every node in the connected graph(including itself).
+-   and only connected to the nodes in the connected graph.
+
+Take example 1 to explain：
+
+-   Node 5's connected node is itself, so it is a connected graph.
+-   Node 0's connected 0, 1, 2. Same as nodes 1, 2.
+-   Nodes 3 and 4 also include themselves and each other.
+
+<!-- tabs:start -->
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int countCompleteComponents(int n, vector<vector<int>>& edges) {
+        int ans = 0;
+        vector<set<int>> m(n + 1, set<int>());
+        for (int i = 0; i < n; i++) {
+            m[i].insert(i);
+        }
+        for (auto x : edges) {
+            m[x[0]].insert(x[1]);
+            m[x[1]].insert(x[0]);
+        }
+        map<set<int>, int> s;
+        for (int i = 0; i < n; i++) {
+            s[m[i]] ++;
+        }
+        for (auto &[x,y] : s) {
+            if (y == x.size()) {
+                ans ++;
+            }
+        }
+        return ans;
+    }
+};
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->
diff --git a/solution/2600-2699/2685.Count the Number of Complete Components/Solution2.cpp b/solution/2600-2699/2685.Count the Number of Complete Components/Solution2.cpp
new file mode 100644
index 0000000000000..1d9762826d9f4
--- /dev/null
+++ b/solution/2600-2699/2685.Count the Number of Complete Components/Solution2.cpp	
@@ -0,0 +1,24 @@
+class Solution {
+public:
+    int countCompleteComponents(int n, vector<vector<int>>& edges) {
+        int ans = 0;
+        vector<set<int>> m(n + 1, set<int>());
+        for (int i = 0; i < n; i++) {
+            m[i].insert(i);
+        }
+        for (auto x : edges) {
+            m[x[0]].insert(x[1]);
+            m[x[1]].insert(x[0]);
+        }
+        map<set<int>, int> s;
+        for (int i = 0; i < n; i++) {
+            s[m[i]] ++;
+        }
+        for (auto &[x,y] : s) {
+            if (y == x.size()) {
+                ans ++;
+            }
+        }
+        return ans;
+    }
+};