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feat: add solutions to lc problem: No.0233 #3539

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Sep 20, 2024
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feat: add solutions to lc problem: No.0233 #3539

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5223b63
feat: add solutions to lc problem: No.0233
yanglbme Sep 20, 2024
3fe17fa
fix: update
yanglbme Sep 20, 2024
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229 changes: 127 additions & 102 deletions solution/0200-0299/0233.Number of Digit One/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -66,18 +66,22 @@ $$

基本步骤如下：

1. 将数字 $n$ 转为 int 数组 $a$，其中 $a[1]$ 为最低位，而 $a[len]$ 为最高位；
1. 根据题目信息，设计函数 $dfs()$，对于本题，我们定义 $dfs(pos, cnt, limit)$，答案为 $dfs(len, 0, true)$。
我们首先将数字 $n$ 转化为字符串 $s$。然后我们设计一个函数 $\textit{dfs}(i, \textit{cnt}, \textit{limit})$，其中：

其中：
- 数字 $i$ 表示当前搜索到的位置，我们从高位开始搜索，即 $i = 0$ 表示最高位。
- 数字 $\textit{cnt}$ 表示当前数字中 $1$ 出现的次数。
- 布尔值 $\textit{limit}$ 表示当前是否受到上界的限制。

- `pos` 表示数字的位数，从末位或者第一位开始，一般根据题目的数字构造性质来选择顺序。对于本题，我们选择从高位开始，因此，`pos` 的初始值为 `len`；
- `cnt` 表示当前数字中包含的 $1$ 的个数。
- `limit` 表示可填的数字的限制，如果无限制，那么可以选择 $[0,1,..9]$，否则，只能选择 $[0,..a[pos]]$。如果 `limit` 为 `true` 且已经取到了能取到的最大值，那么下一个 `limit` 同样为 `true`；如果 `limit` 为 `true` 但是还没有取到最大值，或者 `limit` 为 `false`，那么下一个 `limit` 为 `false`。
函数的执行过程如下：

关于函数的实现细节，可以参考下面的代码。
如果 $i$ 超过了数字 $n$ 的长度，说明搜索结束，直接返回 $cnt$。如果 $\textit{limit}$ 为真，那么 $up$ 为当前数字的第 $i$ 位，否则 $up = 9$。接下来，我们遍历 $j$ 从 $0$ 到 $up$，对于每一个 $j$：

时间复杂度 $O(\log n)$。
- 如果 $j$ 等于 $1$，我们将 $cnt$ 加一。
- 递归调用 $\textit{dfs}(i + 1, \textit{cnt}, \textit{limit} \land j = up)$。

答案为 $\textit{dfs}(0, 0, \text{True})$。

时间复杂度 $O(m^2 \times D)$，空间复杂度 $O(m^2)$。其中 $m$ 为数字 $n$ 的长度，而 $D = 10$。

相似题目：

Expand All @@ -96,57 +100,48 @@ $$
class Solution:
def countDigitOne(self, n: int) -> int:
@cache
def dfs(pos, cnt, limit):
if pos <= 0:
def dfs(i: int, cnt: int, limit: bool) -> int:
if i >= len(s):
return cnt
up = a[pos] if limit else 9
up = int(s[i]) if limit else 9
ans = 0
for i in range(up + 1):
ans += dfs(pos - 1, cnt + (i == 1), limit and i == up)
for j in range(up + 1):
ans += dfs(i + 1, cnt + (j == 1), limit and j == up)
return ans

a = [0] * 12
l = 1
while n:
a[l] = n % 10
n //= 10
l += 1
return dfs(l, 0, True)
s = str(n)
return dfs(0, 0, True)
```

#### Java

```java
class Solution {
private int[] a = new int[12];
private int[][] dp = new int[12][12];
private int m;
private char[] s;
private Integer[][] f;

public int countDigitOne(int n) {
int len = 0;
while (n > 0) {
a[++len] = n % 10;
n /= 10;
}
for (var e : dp) {
Arrays.fill(e, -1);
}
return dfs(len, 0, true);
s = String.valueOf(n).toCharArray();
m = s.length;
f = new Integer[m][m];
return dfs(0, 0, true);
}

private int dfs(int pos, int cnt, boolean limit) {
if (pos <= 0) {
private int dfs(int i, int cnt, boolean limit) {
if (i >= m) {
return cnt;
}
if (!limit && dp[pos][cnt] != -1) {
return dp[pos][cnt];
if (!limit && f[i][cnt] != null) {
return f[i][cnt];
}
int up = limit ? a[pos] : 9;
int up = limit ? s[i] - '0' : 9;
int ans = 0;
for (int i = 0; i <= up; ++i) {
ans += dfs(pos - 1, cnt + (i == 1 ? 1 : 0), limit && i == up);
for (int j = 0; j <= up; ++j) {
ans += dfs(i + 1, cnt + (j == 1 ? 1 : 0), limit && j == up);
}
if (!limit) {
dp[pos][cnt] = ans;
f[i][cnt] = ans;
}
return ans;
}
Expand All @@ -158,35 +153,29 @@ class Solution {
```cpp
class Solution {
public:
int a[12];
int dp[12][12];

int countDigitOne(int n) {
int len = 0;
while (n) {
a[++len] = n % 10;
n /= 10;
}
memset(dp, -1, sizeof dp);
return dfs(len, 0, true);
}

int dfs(int pos, int cnt, bool limit) {
if (pos <= 0) {
return cnt;
}
if (!limit && dp[pos][cnt] != -1) {
return dp[pos][cnt];
}
int ans = 0;
int up = limit ? a[pos] : 9;
for (int i = 0; i <= up; ++i) {
ans += dfs(pos - 1, cnt + (i == 1), limit && i == up);
}
if (!limit) {
dp[pos][cnt] = ans;
}
return ans;
string s = to_string(n);
int m = s.size();
int f[m][m];
memset(f, -1, sizeof(f));
auto dfs = [&](auto&& dfs, int i, int cnt, bool limit) -> int {
if (i >= m) {
return cnt;
}
if (!limit && f[i][cnt] != -1) {
return f[i][cnt];
}
int up = limit ? s[i] - '0' : 9;
int ans = 0;
for (int j = 0; j <= up; ++j) {
ans += dfs(dfs, i + 1, cnt + (j == 1), limit && j == up);
}
if (!limit) {
f[i][cnt] = ans;
}
return ans;
};
return dfs(dfs, 0, 0, true);
}
};
```
Expand All @@ -195,67 +184,103 @@ public:

```go
func countDigitOne(n int) int {
a := make([]int, 12)
dp := make([][]int, 12)
for i := range dp {
dp[i] = make([]int, 12)
for j := range dp[i] {
dp[i][j] = -1
s := strconv.Itoa(n)
m := len(s)
f := make([][]int, m)
for i := range f {
f[i] = make([]int, m)
for j := range f[i] {
f[i][j] = -1
}
}
l := 0
for n > 0 {
l++
a[l] = n % 10
n /= 10
}
var dfs func(int, int, bool) int
dfs = func(pos, cnt int, limit bool) int {
if pos <= 0 {
var dfs func(i, cnt int, limit bool) int
dfs = func(i, cnt int, limit bool) int {
if i >= m {
return cnt
}
if !limit && dp[pos][cnt] != -1 {
return dp[pos][cnt]
if !limit && f[i][cnt] != -1 {
return f[i][cnt]
}
up := 9
if limit {
up = a[pos]
up = int(s[i] - '0')
}
ans := 0
for i := 0; i <= up; i++ {
t := cnt
if i == 1 {
t++
for j := 0; j <= up; j++ {
t := 0
if j == 1 {
t = 1
}
ans += dfs(pos-1, t, limit && i == up)
ans += dfs(i+1, cnt+t, limit && j == up)
}
if !limit {
dp[pos][cnt] = ans
f[i][cnt] = ans
}
return ans
}
return dfs(l, 0, true)
return dfs(0, 0, true)
}
```

#### TypeScript

```ts
function countDigitOne(n: number): number {
const s = n.toString();
const m = s.length;
const f: number[][] = Array.from({ length: m }, () => Array(m).fill(-1));
const dfs = (i: number, cnt: number, limit: boolean): number => {
if (i >= m) {
return cnt;
}
if (!limit && f[i][cnt] !== -1) {
return f[i][cnt];
}
const up = limit ? +s[i] : 9;
let ans = 0;
for (let j = 0; j <= up; ++j) {
ans += dfs(i + 1, cnt + (j === 1 ? 1 : 0), limit && j === up);
}
if (!limit) {
f[i][cnt] = ans;
}
return ans;
};
return dfs(0, 0, true);
}
```

#### C#

```cs
public class Solution {
private int m;
private char[] s;
private int?[,] f;

public int CountDigitOne(int n) {
if (n <= 0) return 0;
if (n < 10) return 1;
return CountDigitOne(n / 10 - 1) * 10 + n / 10 + CountDigitOneOfN(n / 10) * (n % 10 + 1) + (n % 10 >= 1 ? 1 : 0);
s = n.ToString().ToCharArray();
m = s.Length;
f = new int?[m, m];
return Dfs(0, 0, true);
}

private int CountDigitOneOfN(int n) {
var count = 0;
while (n > 0)
{
if (n % 10 == 1) ++count;
n /= 10;
private int Dfs(int i, int cnt, bool limit) {
if (i >= m) {
return cnt;
}
return count;
if (!limit && f[i, cnt] != null) {
return f[i, cnt].Value;
}
int up = limit ? s[i] - '0' : 9;
int ans = 0;
for (int j = 0; j <= up; ++j) {
ans += Dfs(i + 1, cnt + (j == 1 ? 1 : 0), limit && j == up);
}
if (!limit) {
f[i, cnt] = ans;
}
return ans;
}
}
```
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