diff --git a/solution/0800-0899/0822.Card Flipping Game/README.md b/solution/0800-0899/0822.Card Flipping Game/README.md
index 916b116f05012..460459ddaa715 100644
--- a/solution/0800-0899/0822.Card Flipping Game/README.md	
+++ b/solution/0800-0899/0822.Card Flipping Game/README.md	
@@ -66,7 +66,7 @@ tags:
 
 ### 方法一：哈希表
 
-我们注意到，对于位置 $i$，若 $fronts[i]$ 与 $backs[i]$ 元素相同，则一定不满足条件。
+我们注意到，对于位置 $i$，若 $\textit{fronts}[i]$ 与 $\textit{backs}[i]$ 元素相同，则一定不满足条件。
 
 因此，我们先找出正面与背面相同的元素，记录在哈希表 $s$ 中。
 
@@ -195,6 +195,43 @@ function flipgame(fronts: number[], backs: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashSet;
+
+impl Solution {
+    pub fn flipgame(fronts: Vec<i32>, backs: Vec<i32>) -> i32 {
+        let n = fronts.len();
+        let mut s: HashSet<i32> = HashSet::new();
+
+        for i in 0..n {
+            if fronts[i] == backs[i] {
+                s.insert(fronts[i]);
+            }
+        }
+
+        let mut ans = 9999;
+        for &v in fronts.iter() {
+            if !s.contains(&v) {
+                ans = ans.min(v);
+            }
+        }
+        for &v in backs.iter() {
+            if !s.contains(&v) {
+                ans = ans.min(v);
+            }
+        }
+
+        if ans == 9999 {
+            0
+        } else {
+            ans
+        }
+    }
+}
+```
+
 #### C#
 
 ```cs
diff --git a/solution/0800-0899/0822.Card Flipping Game/README_EN.md b/solution/0800-0899/0822.Card Flipping Game/README_EN.md
index 89abfd94a2f80..dca0480e715a6 100644
--- a/solution/0800-0899/0822.Card Flipping Game/README_EN.md	
+++ b/solution/0800-0899/0822.Card Flipping Game/README_EN.md	
@@ -59,7 +59,17 @@ There are no good integers no matter how we flip the cards, so we return 0.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Hash Table
+
+We observe that for position $i$, if $\textit{fronts}[i]$ is equal to $\textit{backs}[i]$, then it certainly does not satisfy the condition.
+
+Therefore, we first identify all elements that appear the same on both the front and back sides and record them in a hash set $s$.
+
+Next, we iterate through all elements in both the front and back arrays. For any element $x$ that is **not** in the hash set $s$, we update the minimum value of the answer.
+
+Finally, if we find any element that satisfies the condition, we return the minimum answer; otherwise, we return $0$.
+
+The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the arrays.
 
 <!-- tabs:start -->
 
@@ -180,6 +190,43 @@ function flipgame(fronts: number[], backs: number[]): number {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashSet;
+
+impl Solution {
+    pub fn flipgame(fronts: Vec<i32>, backs: Vec<i32>) -> i32 {
+        let n = fronts.len();
+        let mut s: HashSet<i32> = HashSet::new();
+
+        for i in 0..n {
+            if fronts[i] == backs[i] {
+                s.insert(fronts[i]);
+            }
+        }
+
+        let mut ans = 9999;
+        for &v in fronts.iter() {
+            if !s.contains(&v) {
+                ans = ans.min(v);
+            }
+        }
+        for &v in backs.iter() {
+            if !s.contains(&v) {
+                ans = ans.min(v);
+            }
+        }
+
+        if ans == 9999 {
+            0
+        } else {
+            ans
+        }
+    }
+}
+```
+
 #### C#
 
 ```cs
diff --git a/solution/0800-0899/0822.Card Flipping Game/Solution.rs b/solution/0800-0899/0822.Card Flipping Game/Solution.rs
new file mode 100644
index 0000000000000..25f15912f0308
--- /dev/null
+++ b/solution/0800-0899/0822.Card Flipping Game/Solution.rs	
@@ -0,0 +1,32 @@
+use std::collections::HashSet;
+
+impl Solution {
+    pub fn flipgame(fronts: Vec<i32>, backs: Vec<i32>) -> i32 {
+        let n = fronts.len();
+        let mut s: HashSet<i32> = HashSet::new();
+
+        for i in 0..n {
+            if fronts[i] == backs[i] {
+                s.insert(fronts[i]);
+            }
+        }
+
+        let mut ans = 9999;
+        for &v in fronts.iter() {
+            if !s.contains(&v) {
+                ans = ans.min(v);
+            }
+        }
+        for &v in backs.iter() {
+            if !s.contains(&v) {
+                ans = ans.min(v);
+            }
+        }
+
+        if ans == 9999 {
+            0
+        } else {
+            ans
+        }
+    }
+}