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Commit 1331dbb

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feat: add solutions to lc problem: No.0400.Nth Digit
1 parent 06875ce commit 1331dbb

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4 files changed

+114
-28
lines changed

4 files changed

+114
-28
lines changed

‎solution/0400-0499/0400.Nth Digit/README.md‎

Lines changed: 46 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -29,7 +29,6 @@
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<strong>解释:</strong>第 11 位数字在序列 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... 里是 <strong>0 </strong>,它是 10 的一部分。
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</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
@@ -41,15 +40,60 @@
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
44-
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class Solution:
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def findNthDigit(self, n: int) -> int:
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bits, t = 1, 9
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while n > bits * t:
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n -= bits * t
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bits += 1
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t *= 10
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start = 10 ** (bits - 1) + (n // bits) - 1
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if n % bits == 0:
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return start % 10
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return int(str((start + 1))[(n % bits) - 1])
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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 findNthDigit(int n) {
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int bits = 1, t = 9;
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while (n / bits > t) {
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n -= bits * t;
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++bits;
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t *= 10;
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}
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int start = (int) Math.pow(10, bits - 1) + (n / bits) - 1;
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if (n % bits == 0) {
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return start % 10;
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}
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return String.valueOf(start + 1).charAt((n % bits) - 1) - '0';
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}
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}
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```
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79+
### **C++**
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```cpp
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class Solution {
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public:
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int findNthDigit(int n) {
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int bits = 1, t = 9;
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while (n / bits > t)
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{
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n -= bits * t;
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++bits;
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t *= 10;
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}
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int start = pow(10, bits - 1) + (n / bits) - 1;
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if (n % bits == 0) return start % 10;
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return to_string(start + 1)[(n % bits) - 1] - '0';
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}
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};
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```
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### **...**

‎solution/0400-0499/0400.Nth Digit/README_EN.md‎

Lines changed: 46 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -29,21 +29,65 @@
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<li><code>1 &lt;= n &lt;= 2<sup>31</sup> - 1</code></li>
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</ul>
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32-
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def findNthDigit(self, n: int) -> int:
41+
bits, t = 1, 9
42+
while n > bits * t:
43+
n -= bits * t
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bits += 1
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t *= 10
46+
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start = 10 ** (bits - 1) + (n // bits) - 1
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if n % bits == 0:
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return start % 10
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return int(str((start + 1))[(n % bits) - 1])
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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 findNthDigit(int n) {
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int bits = 1, t = 9;
59+
while (n / bits > t) {
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n -= bits * t;
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++bits;
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t *= 10;
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}
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int start = (int) Math.pow(10, bits - 1) + (n / bits) - 1;
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if (n % bits == 0) {
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return start % 10;
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}
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return String.valueOf(start + 1).charAt((n % bits) - 1) - '0';
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int findNthDigit(int n) {
79+
int bits = 1, t = 9;
80+
while (n / bits > t)
81+
{
82+
n -= bits * t;
83+
++bits;
84+
t *= 10;
85+
}
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int start = pow(10, bits - 1) + (n / bits) - 1;
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if (n % bits == 0) return start % 10;
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return to_string(start + 1)[(n % bits) - 1] - '0';
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}
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};
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```
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### **...**
Lines changed: 10 additions & 24 deletions
Original file line numberDiff line numberDiff line change
@@ -1,29 +1,15 @@
11
class Solution {
2-
/***
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* 12345678910111213
4-
* 规律个位数9个数一共有9*1,两位数90个数 一共有90*2个数字,三位数有900个数一共有900*3个数字,以此类推
5-
* 举例15,15-9=6,6/2=3...0,余数是0,那么这个数值value=10*(2-1)+(3-1)=12,整除取最后一位 12%10=2
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* 举例14,14-9=5,5/2=2...1,余数不为0,那么这个数值value=10*(2-1)+2=12,则为这个数的第余数个 12/(10*(2-1))%10=1
7-
*/
82
public int findNthDigit(int n) {
9-
long max = 9;
10-
long num = n;
11-
long digits = 1;
12-
while (n > 0) {
13-
if (num - max * digits > 0) {
14-
num = num - max * digits;
15-
digits++;
16-
max = max * 10;
17-
} else {
18-
long count = num / digits;
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long childDigits = num % digits;
20-
if (childDigits == 0) {
21-
return (int) (((long) Math.pow(10, digits - 1) + count - 1) % 10);
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} else {
23-
return (int) (((long) Math.pow(10, digits - 1) + count) / ((long) Math.pow(10, (digits - childDigits))) % 10);
24-
}
25-
}
3+
int bits = 1, t = 9;
4+
while (n / bits > t) {
5+
n -= bits * t;
6+
++bits;
7+
t *= 10;
268
}
27-
return 0;
9+
int start = (int) Math.pow(10, bits - 1) + (n / bits) - 1;
10+
if (n % bits == 0) {
11+
return start % 10;
12+
}
13+
return String.valueOf(start + 1).charAt((n % bits) - 1) - '0';
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}
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}
Lines changed: 12 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,12 @@
1+
class Solution:
2+
def findNthDigit(self, n: int) -> int:
3+
bits, t = 1, 9
4+
while n > bits * t:
5+
n -= bits * t
6+
bits += 1
7+
t *= 10
8+
9+
start = 10 ** (bits - 1) + (n // bits) - 1
10+
if n % bits == 0:
11+
return start % 10
12+
return int(str((start + 1))[(n % bits) - 1])

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