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feat: add solutions to lc problem: No.0233.Number of Digit One
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‎solution/0200-0299/0233.Number of Digit One/README.md

+107-1
Original file line numberDiff line numberDiff line change
@@ -37,22 +37,128 @@
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<!-- 这里可写通用的实现逻辑 -->
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经典数位 dp 问题，也可以用找规律解决
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def countDigitOne(self, n: int) -> int:
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dp = [[-1] * 10 for _ in range(10)]
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digit = []
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while n:
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digit.append(n % 10)
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n //= 10
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def dfs(pos: int, cnt: int, limit: bool) -> int:
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if pos == -1: return cnt
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if not limit and dp[pos][cnt] != -1: return dp[pos][cnt]
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up = digit[pos] if limit else 9
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ans = 0
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for i in range(up + 1):
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nxt = cnt + 1 if i == 1 else cnt
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ans += dfs(pos - 1, nxt, limit and i == digit[pos])
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if not limit: dp[pos][cnt] = ans
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return ans
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return dfs(len(digit) - 1, 0, True)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countDigitOne(int n) {
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int index = 1;
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int count = 0;
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int high = n,cur = 0,low = 0;
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while(high > 0){
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high /= 10;
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cur = (n / index) % 10;
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low = n - (n / index) * index;
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if(cur == 0) count += high * index;
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if(cur == 1) count += high * index + low + 1;
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if(cur > 1) count += (high+1) * index;
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index *= 10;
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}
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return count;
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}
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}
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```
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### **C#**
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```cs
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public class Solution {
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public int CountDigitOne(int n) {
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if (n <= 0) return 0;
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if (n < 10) return 1;
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return CountDigitOne(n / 10 - 1) * 10 + n / 10 + CountDigitOneOfN(n / 10) * (n % 10 + 1) + (n % 10 >= 1 ? 1 : 0);
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}
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private int CountDigitOneOfN(int n) {
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var count = 0;
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while (n > 0)
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{
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if (n % 10 == 1) ++count;
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n /= 10;
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}
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return count;
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}
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}
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```
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### **Go**
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```go
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func countDigitOne(n int) int {
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digit := make([]int, 0)
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for i := n; i > 0; i /= 10 {
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digit = append(digit, i%10)
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}
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dp := make([][]int, 10)
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for i := range dp {
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dp[i] = make([]int, 10)
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for j := 0; j < 10; j++ {
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dp[i][j] = -1
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}
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}
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var dfs func(pos, cnt int, limit bool) int
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dfs = func(pos, cnt int, limit bool) int {
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if pos == -1 {
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return cnt
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}
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if !limit && dp[pos][cnt] != -1 {
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return dp[pos][cnt]
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}
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up := 9
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if limit {
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up = digit[pos]
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}
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ans := 0
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for i := 0; i <= up; i++ {
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next := cnt
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if i == 1 {
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next++
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}
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ans += dfs(pos-1, next, limit && i == digit[pos])
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}
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if !limit {
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dp[pos][cnt] = ans
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}
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return ans
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}
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return dfs(len(digit)-1, 0, true)
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}
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```
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### **...**

‎solution/0200-0299/0233.Number of Digit One/README_EN.md

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@@ -31,18 +31,124 @@
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## Solutions
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simple digital dp problem (or it can be solved by finding a rule)
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<!-- tabs:start -->
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### **Python3**
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```python
39-
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class Solution:
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def countDigitOne(self, n: int) -> int:
43+
dp = [[-1] * 10 for _ in range(10)]
44+
digit = []
45+
while n:
46+
digit.append(n % 10)
47+
n //= 10
48+
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def dfs(pos: int, cnt: int, limit: bool) -> int:
50+
if pos == -1: return cnt
51+
if not limit and dp[pos][cnt] != -1: return dp[pos][cnt]
52+
up = digit[pos] if limit else 9
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ans = 0
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for i in range(up + 1):
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nxt = cnt + 1 if i == 1 else cnt
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ans += dfs(pos - 1, nxt, limit and i == digit[pos])
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if not limit: dp[pos][cnt] = ans
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return ans
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return dfs(len(digit) - 1, 0, True)
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```
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### **Java**
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```java
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class Solution {
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public int countDigitOne(int n) {
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int index = 1;
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int count = 0;
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int high = n,cur = 0,low = 0;
71+
while(high > 0){
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high /= 10;
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cur = (n / index) % 10;
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low = n - (n / index) * index;
75+
if(cur == 0) count += high * index;
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if(cur == 1) count += high * index + low + 1;
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if(cur > 1) count += (high+1) * index;
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index *= 10;
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}
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return count;
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}
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}
83+
```
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### **C#**
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```cs
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public class Solution {
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public int CountDigitOne(int n) {
90+
if (n <= 0) return 0;
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if (n < 10) return 1;
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return CountDigitOne(n / 10 - 1) * 10 + n / 10 + CountDigitOneOfN(n / 10) * (n % 10 + 1) + (n % 10 >= 1 ? 1 : 0);
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}
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private int CountDigitOneOfN(int n) {
96+
var count = 0;
97+
while (n > 0)
98+
{
99+
if (n % 10 == 1) ++count;
100+
n /= 10;
101+
}
102+
return count;
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}
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}
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```
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### **Go**
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```go
110+
func countDigitOne(n int) int {
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digit := make([]int, 0)
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for i := n; i > 0; i /= 10 {
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digit = append(digit, i%10)
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}
115+
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dp := make([][]int, 10)
117+
for i := range dp {
118+
dp[i] = make([]int, 10)
119+
for j := 0; j < 10; j++ {
120+
dp[i][j] = -1
121+
}
122+
}
123+
124+
var dfs func(pos, cnt int, limit bool) int
125+
dfs = func(pos, cnt int, limit bool) int {
126+
if pos == -1 {
127+
return cnt
128+
}
129+
if !limit && dp[pos][cnt] != -1 {
130+
return dp[pos][cnt]
131+
}
132+
up := 9
133+
if limit {
134+
up = digit[pos]
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}
136+
ans := 0
137+
for i := 0; i <= up; i++ {
138+
next := cnt
139+
if i == 1 {
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next++
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}
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ans += dfs(pos-1, next, limit && i == digit[pos])
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}
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if !limit {
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dp[pos][cnt] = ans
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}
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return ans
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}
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return dfs(len(digit)-1, 0, true)
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}
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```
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48154
### **...**

‎solution/0200-0299/0233.Number of Digit One/Solution.go

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,42 @@
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func countDigitOne(n int) int {
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digit := make([]int, 0)
3+
for i := n; i > 0; i /= 10 {
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digit = append(digit, i%10)
5+
}
6+
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dp := make([][]int, 10)
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for i := range dp {
9+
dp[i] = make([]int, 10)
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for j := 0; j < 10; j++ {
11+
dp[i][j] = -1
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}
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}
14+
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var dfs func(pos, cnt int, limit bool) int
16+
dfs = func(pos, cnt int, limit bool) int {
17+
if pos == -1 {
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return cnt
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}
20+
if !limit && dp[pos][cnt] != -1 {
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return dp[pos][cnt]
22+
}
23+
up := 9
24+
if limit {
25+
up = digit[pos]
26+
}
27+
ans := 0
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for i := 0; i <= up; i++ {
29+
next := cnt
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if i == 1 {
31+
next++
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}
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ans += dfs(pos-1, next, limit && i == digit[pos])
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}
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if !limit {
36+
dp[pos][cnt] = ans
37+
}
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return ans
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}
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return dfs(len(digit)-1, 0, true)
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}

‎solution/0200-0299/0233.Number of Digit One/Solution.py

+20
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
1+
class Solution:
2+
def countDigitOne(self, n: int) -> int:
3+
dp = [[-1] * 10 for _ in range(10)]
4+
digit = []
5+
while n:
6+
digit.append(n % 10)
7+
n //= 10
8+
9+
def dfs(pos: int, cnt: int, limit: bool) -> int:
10+
if pos == -1: return cnt
11+
if not limit and dp[pos][cnt] != -1: return dp[pos][cnt]
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up = digit[pos] if limit else 9
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ans = 0
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for i in range(up + 1):
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nxt = cnt + 1 if i == 1 else cnt
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ans += dfs(pos - 1, nxt, limit and i == digit[pos])
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if not limit: dp[pos][cnt] = ans
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return ans
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return dfs(len(digit) - 1, 0, True)

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