feat: add solutions to lc problem: No.1015

No.1015.Smallest Integer Divisible by K

Author: yanglbme

solution/1000-1099/1015.Smallest Integer Divisible by K/README.md

@@ -47,22 +47,81 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：数学**
+
+我们注意到，正整数 $n$ 初始值为 $1$，每次乘以 $10$ 后再加 $1$，即 $n = n \times 10 + 1$，而 $(n \times 10 + 1) \bmod k = ((n \bmod k) \times 10 + 1) \bmod k$，因此我们可以通过计算 $n \bmod k$ 来判断 $n$ 是否能被 $k$ 整除。
+
+我们从 $n = 1$ 开始，每次计算 $n \bmod k$，直到 $n \bmod k = 0$，此时 $n$ 就是我们要求的最小正整数，其长度即为 $n$ 的位数。否则，我们更新 $n = (n \times 10 + 1) \bmod k$。如果循环 $k$ 次后，仍然没有找到 $n \bmod k = 0$，则说明不存在这样的 $n$，返回 $-1$。
+
+时间复杂度 $O(k)$，空间复杂度 $O(1)$。其中 $k$ 为给定的正整数。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def smallestRepunitDivByK(self, k: int) -> int:
+        n = 1 % k
+        for i in range(1, k + 1):
+            if n == 0:
+                return i
+            n = (n * 10 + 1) % k
+        return -1
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int smallestRepunitDivByK(int k) {
+        int n = 1 % k;
+        for (int i = 1; i <= k; ++i) {
+            if (n == 0) {
+                return i;
+            }
+            n = (n * 10 + 1) % k;
+        }
+        return -1;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int smallestRepunitDivByK(int k) {
+        int n = 1 % k;
+        for (int i = 1; i <= k; ++i) {
+            if (n == 0) {
+                return i;
+            }
+            n = (n * 10 + 1) % k;
+        }
+        return -1;
+    }
+};
+```
 
+### **Go**
+
+```go
+func smallestRepunitDivByK(k int) int {
+	n := 1 % k
+	for i := 1; i <= k; i++ {
+		if n == 0 {
+			return i
+		}
+		n = (n*10 + 1) % k
+	}
+	return -1
+}
 ```
 
 ### **...**

solution/1000-1099/1015.Smallest Integer Divisible by K/README_EN.md

@@ -49,13 +49,64 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def smallestRepunitDivByK(self, k: int) -> int:
+        n = 1 % k
+        for i in range(1, k + 1):
+            if n == 0:
+                return i
+            n = (n * 10 + 1) % k
+        return -1
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int smallestRepunitDivByK(int k) {
+        int n = 1 % k;
+        for (int i = 1; i <= k; ++i) {
+            if (n == 0) {
+                return i;
+            }
+            n = (n * 10 + 1) % k;
+        }
+        return -1;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int smallestRepunitDivByK(int k) {
+        int n = 1 % k;
+        for (int i = 1; i <= k; ++i) {
+            if (n == 0) {
+                return i;
+            }
+            n = (n * 10 + 1) % k;
+        }
+        return -1;
+    }
+};
+```
 
+### **Go**
+
+```go
+func smallestRepunitDivByK(k int) int {
+	n := 1 % k
+	for i := 1; i <= k; i++ {
+		if n == 0 {
+			return i
+		}
+		n = (n*10 + 1) % k
+	}
+	return -1
+}
 ```
 
 ### **...**

solution/1000-1099/1015.Smallest Integer Divisible by K/Solution.cpp

@@ -0,0 +1,13 @@
+class Solution {
+public:
+    int smallestRepunitDivByK(int k) {
+        int n = 1 % k;
+        for (int i = 1; i <= k; ++i) {
+            if (n == 0) {
+                return i;
+            }
+            n = (n * 10 + 1) % k;
+        }
+        return -1;
+    }
+};

solution/1000-1099/1015.Smallest Integer Divisible by K/Solution.go

@@ -0,0 +1,10 @@
+func smallestRepunitDivByK(k int) int {
+	n := 1 % k
+	for i := 1; i <= k; i++ {
+		if n == 0 {
+			return i
+		}
+		n = (n*10 + 1) % k
+	}
+	return -1
+}

solution/1000-1099/1015.Smallest Integer Divisible by K/Solution.java

@@ -0,0 +1,12 @@
+class Solution {
+    public int smallestRepunitDivByK(int k) {
+        int n = 1 % k;
+        for (int i = 1; i <= k; ++i) {
+            if (n == 0) {
+                return i;
+            }
+            n = (n * 10 + 1) % k;
+        }
+        return -1;
+    }
+}

solution/1000-1099/1015.Smallest Integer Divisible by K/Solution.py

@@ -0,0 +1,8 @@
+class Solution:
+    def smallestRepunitDivByK(self, k: int) -> int:
+        n = 1 % k
+        for i in range(1, k + 1):
+            if n == 0:
+                return i
+            n = (n * 10 + 1) % k
+        return -1