feat: add rust solution to lc problem: No.2048

solution/2000-2099/2048.Next Greater Numerically Balanced Number/README.md

@@ -35,7 +35,7 @@ tags:
 <strong>输出：</strong>22
 <strong>解释：</strong>
 22 是一个数值平衡数，因为：
-- 数字 2 出现 2 次 
+- 数字 2 出现 2 次
 这也是严格大于 1 的最小数值平衡数。
 </pre>
 
@@ -47,7 +47,7 @@ tags:
 <strong>解释：</strong>
 1333 是一个数值平衡数，因为：
 - 数字 1 出现 1 次。
-- 数字 3 出现 3 次。 
+- 数字 3 出现 3 次。
 这也是严格大于 1000 的最小数值平衡数。
 注意，1022 不能作为本输入的答案，因为数字 0 的出现次数超过了 0 。</pre>
 
@@ -59,7 +59,7 @@ tags:
 <strong>解释：</strong>
 3133 是一个数值平衡数，因为：
 - 数字 1 出现 1 次。
-- 数字 3 出现 3 次。 
+- 数字 3 出现 3 次。
 这也是严格大于 3000 的最小数值平衡数。
 </pre>
 
@@ -197,6 +197,38 @@ function nextBeautifulNumber(n: number): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn next_beautiful_number(n: i32) -> i32 {
+        let mut x = n + 1;
+        loop {
+            let mut cnt = [0; 10];
+            let mut y = x;
+            while y > 0 {
+                cnt[(y % 10) as usize] += 1;
+                y /= 10;
+            }
+            let mut ok = true;
+            let mut y2 = x;
+            while y2 > 0 {
+                let d = (y2 % 10) as usize;
+                if d != cnt[d] {
+                    ok = false;
+                    break;
+                }
+                y2 /= 10;
+            }
+            if ok {
+                return x;
+            }
+            x += 1;
+        }
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2000-2099/2048.Next Greater Numerically Balanced Number/README_EN.md

@@ -32,9 +32,9 @@ tags:
 <pre>
 <strong>Input:</strong> n = 1
 <strong>Output:</strong> 22
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 22 is numerically balanced since:
-- The digit 2 occurs 2 times. 
+- The digit 2 occurs 2 times.
 It is also the smallest numerically balanced number strictly greater than 1.
 </pre>
 
@@ -43,10 +43,10 @@ It is also the smallest numerically balanced number strictly greater than 1.
 <pre>
 <strong>Input:</strong> n = 1000
 <strong>Output:</strong> 1333
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 1333 is numerically balanced since:
 - The digit 1 occurs 1 time.
-- The digit 3 occurs 3 times. 
+- The digit 3 occurs 3 times.
 It is also the smallest numerically balanced number strictly greater than 1000.
 Note that 1022 cannot be the answer because 0 appeared more than 0 times.
 </pre>
@@ -56,7 +56,7 @@ Note that 1022 cannot be the answer because 0 appeared more than 0 times.
 <pre>
 <strong>Input:</strong> n = 3000
 <strong>Output:</strong> 3133
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 3133 is numerically balanced since:
 - The digit 1 occurs 1 time.
 - The digit 3 occurs 3 times.
@@ -196,6 +196,38 @@ function nextBeautifulNumber(n: number): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn next_beautiful_number(n: i32) -> i32 {
+        let mut x = n + 1;
+        loop {
+            let mut cnt = [0; 10];
+            let mut y = x;
+            while y > 0 {
+                cnt[(y % 10) as usize] += 1;
+                y /= 10;
+            }
+            let mut ok = true;
+            let mut y2 = x;
+            while y2 > 0 {
+                let d = (y2 % 10) as usize;
+                if d != cnt[d] {
+                    ok = false;
+                    break;
+                }
+                y2 /= 10;
+            }
+            if ok {
+                return x;
+            }
+            x += 1;
+        }
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2000-2099/2048.Next Greater Numerically Balanced Number/Solution.rs

@@ -0,0 +1,27 @@
+impl Solution {
+    pub fn next_beautiful_number(n: i32) -> i32 {
+        let mut x = n + 1;
+        loop {
+            let mut cnt = [0; 10];
+            let mut y = x;
+            while y > 0 {
+                cnt[(y % 10) as usize] += 1;
+                y /= 10;
+            }
+            let mut ok = true;
+            let mut y2 = x;
+            while y2 > 0 {
+                let d = (y2 % 10) as usize;
+                if d != cnt[d] {
+                    ok = false;
+                    break;
+                }
+                y2 /= 10;
+            }
+            if ok {
+                return x;
+            }
+            x += 1;
+        }
+    }
+}