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Commit 6445c81

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feat: add rust solution to lc problem: No.2048 (#4796)
1 parent 7ffcdf6 commit 6445c81

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3 files changed

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-8
lines changed

3 files changed

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

Lines changed: 35 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -35,7 +35,7 @@ tags:
3535
<strong>输出:</strong>22
3636
<strong>解释:</strong>
3737
22 是一个数值平衡数,因为:
38-
- 数字 2 出现 2 次
38+
- 数字 2 出现 2 次
3939
这也是严格大于 1 的最小数值平衡数。
4040
</pre>
4141

@@ -47,7 +47,7 @@ tags:
4747
<strong>解释:</strong>
4848
1333 是一个数值平衡数,因为:
4949
- 数字 1 出现 1 次。
50-
- 数字 3 出现 3 次。
50+
- 数字 3 出现 3 次。
5151
这也是严格大于 1000 的最小数值平衡数。
5252
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>
5353

@@ -59,7 +59,7 @@ tags:
5959
<strong>解释:</strong>
6060
3133 是一个数值平衡数,因为:
6161
- 数字 1 出现 1 次。
62-
- 数字 3 出现 3 次。
62+
- 数字 3 出现 3 次。
6363
这也是严格大于 3000 的最小数值平衡数。
6464
</pre>
6565

@@ -197,6 +197,38 @@ function nextBeautifulNumber(n: number): number {
197197
}
198198
```
199199

200+
#### Rust
201+
202+
```rust
203+
impl Solution {
204+
pub fn next_beautiful_number(n: i32) -> i32 {
205+
let mut x = n + 1;
206+
loop {
207+
let mut cnt = [0; 10];
208+
let mut y = x;
209+
while y > 0 {
210+
cnt[(y % 10) as usize] += 1;
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y /= 10;
212+
}
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let mut ok = true;
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let mut y2 = x;
215+
while y2 > 0 {
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let d = (y2 % 10) as usize;
217+
if d != cnt[d] {
218+
ok = false;
219+
break;
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}
221+
y2 /= 10;
222+
}
223+
if ok {
224+
return x;
225+
}
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x += 1;
227+
}
228+
}
229+
}
230+
```
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200232
<!-- tabs:end -->
201233

202234
<!-- solution:end -->

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

Lines changed: 37 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -32,9 +32,9 @@ tags:
3232
<pre>
3333
<strong>Input:</strong> n = 1
3434
<strong>Output:</strong> 22
35-
<strong>Explanation:</strong>
35+
<strong>Explanation:</strong>
3636
22 is numerically balanced since:
37-
- The digit 2 occurs 2 times.
37+
- The digit 2 occurs 2 times.
3838
It is also the smallest numerically balanced number strictly greater than 1.
3939
</pre>
4040

@@ -43,10 +43,10 @@ It is also the smallest numerically balanced number strictly greater than 1.
4343
<pre>
4444
<strong>Input:</strong> n = 1000
4545
<strong>Output:</strong> 1333
46-
<strong>Explanation:</strong>
46+
<strong>Explanation:</strong>
4747
1333 is numerically balanced since:
4848
- The digit 1 occurs 1 time.
49-
- The digit 3 occurs 3 times.
49+
- The digit 3 occurs 3 times.
5050
It is also the smallest numerically balanced number strictly greater than 1000.
5151
Note that 1022 cannot be the answer because 0 appeared more than 0 times.
5252
</pre>
@@ -56,7 +56,7 @@ Note that 1022 cannot be the answer because 0 appeared more than 0 times.
5656
<pre>
5757
<strong>Input:</strong> n = 3000
5858
<strong>Output:</strong> 3133
59-
<strong>Explanation:</strong>
59+
<strong>Explanation:</strong>
6060
3133 is numerically balanced since:
6161
- The digit 1 occurs 1 time.
6262
- The digit 3 occurs 3 times.
@@ -196,6 +196,38 @@ function nextBeautifulNumber(n: number): number {
196196
}
197197
```
198198

199+
#### Rust
200+
201+
```rust
202+
impl Solution {
203+
pub fn next_beautiful_number(n: i32) -> i32 {
204+
let mut x = n + 1;
205+
loop {
206+
let mut cnt = [0; 10];
207+
let mut y = x;
208+
while y > 0 {
209+
cnt[(y % 10) as usize] += 1;
210+
y /= 10;
211+
}
212+
let mut ok = true;
213+
let mut y2 = x;
214+
while y2 > 0 {
215+
let d = (y2 % 10) as usize;
216+
if d != cnt[d] {
217+
ok = false;
218+
break;
219+
}
220+
y2 /= 10;
221+
}
222+
if ok {
223+
return x;
224+
}
225+
x += 1;
226+
}
227+
}
228+
}
229+
```
230+
199231
<!-- tabs:end -->
200232

201233
<!-- solution:end -->
Lines changed: 27 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,27 @@
1+
impl Solution {
2+
pub fn next_beautiful_number(n: i32) -> i32 {
3+
let mut x = n + 1;
4+
loop {
5+
let mut cnt = [0; 10];
6+
let mut y = x;
7+
while y > 0 {
8+
cnt[(y % 10) as usize] += 1;
9+
y /= 10;
10+
}
11+
let mut ok = true;
12+
let mut y2 = x;
13+
while y2 > 0 {
14+
let d = (y2 % 10) as usize;
15+
if d != cnt[d] {
16+
ok = false;
17+
break;
18+
}
19+
y2 /= 10;
20+
}
21+
if ok {
22+
return x;
23+
}
24+
x += 1;
25+
}
26+
}
27+
}

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