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feat: add solutions to lc problem: No.1492
No.1492.The kth Factor of n
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‎solution/1400-1499/1492.The kth Factor of n/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：暴力枚举**
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“因子”是指能整除某个数的数。因此，我们只需要从小到大枚举 $[1,2,..n]$，找到所有能整除 $n$ 的数，然后返回第 $k$ 个即可。
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时间复杂度 $O(n)$。
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**方法二：枚举优化**
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我们可以发现，如果 $n$ 有一个因子 $x$，那么 $n$ 一定也有一个因子 $n/x$。
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因此，我们先需要枚举 $[1,2,...\left \lfloor \sqrt{n} \right \rfloor ]$，找到所有能整除 $n$ 的数，如果找到第 $k$ 个因子，那么直接返回即可。如果没有找到第 $k$ 个因子，那么我们再倒序枚举 $[\left \lfloor \sqrt{n} \right \rfloor ,..1]$，找到第 $k$ 个因子即可。
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时间复杂度 $O(\sqrt{n})$。
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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 kthFactor(self, n: int, k: int) -> int:
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for i in range(1, n + 1):
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if n % i == 0:
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k -= 1
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if k == 0:
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return i
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return -1
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```
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```python
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class Solution:
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def kthFactor(self, n: int, k: int) -> int:
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i = 1
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while i * i < n:
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if n % i == 0:
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k -= 1
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if k == 0:
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return i
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i += 1
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if i * i != n:
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i -= 1
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while i:
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if (n % (n // i)) == 0:
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k -= 1
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if k == 0:
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return n // i
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i -= 1
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return -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 kthFactor(int n, int k) {
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for (int i = 1; i <= n; ++i) {
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if (n % i == 0 && (--k == 0)) {
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return i;
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}
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}
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return -1;
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}
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}
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```
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```java
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class Solution {
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public int kthFactor(int n, int k) {
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int i = 1;
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for (; i < n / i; ++i) {
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if (n % i == 0 && (--k == 0)) {
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return i;
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}
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}
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if (i * i != n) {
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--i;
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}
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for (; i > 0; --i) {
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if (n % (n / i) == 0 && (--k == 0)) {
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return n / i;
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}
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}
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return -1;
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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 kthFactor(int n, int k) {
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int i = 1;
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for (; i < n / i; ++i) {
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if (n % i == 0 && (--k == 0)) {
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return i;
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}
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}
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if (i * i != n) {
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--i;
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}
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for (; i > 0; --i) {
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if (n % (n / i) == 0 && (--k == 0)) {
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return n / i;
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}
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}
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return -1;
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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 kthFactor(n int, k int) int {
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for i := 1; i <= n; i++ {
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if n%i == 0 {
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k--
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if k == 0 {
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return i
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}
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}
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}
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return -1
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}
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```
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```go
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func kthFactor(n int, k int) int {
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i := 1
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for ; i < n/i; i++ {
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if n%i == 0 {
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k--
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if k == 0 {
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return i
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}
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}
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}
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if i*i != n {
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i--
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}
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for ; i > 0; i-- {
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if n%(n/i) == 0 {
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k--
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if k == 0 {
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return n / i
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}
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}
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}
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return -1
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}
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```
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### **...**

‎solution/1400-1499/1492.The kth Factor of n/README_EN.md

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### **Python3**
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```python
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class Solution:
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def kthFactor(self, n: int, k: int) -> int:
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for i in range(1, n + 1):
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if n % i == 0:
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k -= 1
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if k == 0:
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return i
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return -1
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```
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```python
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class Solution:
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def kthFactor(self, n: int, k: int) -> int:
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i = 1
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while i * i < n:
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if n % i == 0:
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k -= 1
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if k == 0:
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return i
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i += 1
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if i * i != n:
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i -= 1
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while i:
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if (n % (n // i)) == 0:
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k -= 1
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if k == 0:
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return n // i
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i -= 1
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return -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 kthFactor(int n, int k) {
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for (int i = 1; i <= n; ++i) {
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if (n % i == 0 && (--k == 0)) {
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return i;
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}
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}
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return -1;
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}
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}
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```
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```java
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class Solution {
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public int kthFactor(int n, int k) {
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int i = 1;
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for (; i < n / i; ++i) {
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if (n % i == 0 && (--k == 0)) {
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return i;
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}
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}
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if (i * i != n) {
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--i;
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}
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for (; i > 0; --i) {
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if (n % (n / i) == 0 && (--k == 0)) {
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return n / i;
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}
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}
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return -1;
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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 kthFactor(int n, int k) {
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int i = 1;
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for (; i < n / i; ++i) {
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if (n % i == 0 && (--k == 0)) {
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return i;
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}
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}
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if (i * i != n) {
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--i;
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}
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for (; i > 0; --i) {
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if (n % (n / i) == 0 && (--k == 0)) {
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return n / i;
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}
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}
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return -1;
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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 kthFactor(n int, k int) int {
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for i := 1; i <= n; i++ {
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if n%i == 0 {
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k--
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if k == 0 {
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return i
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}
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}
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}
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return -1
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}
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```
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```go
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func kthFactor(n int, k int) int {
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i := 1
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for ; i < n/i; i++ {
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if n%i == 0 {
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k--
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if k == 0 {
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return i
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}
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}
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}
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if i*i != n {
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i--
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}
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for ; i > 0; i-- {
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if n%(n/i) == 0 {
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k--
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if k == 0 {
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return n / i
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}
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}
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}
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return -1
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}
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```
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### **...**

‎solution/1400-1499/1492.The kth Factor of n/Solution.cpp

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@@ -0,0 +1,20 @@
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class Solution {
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public:
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int kthFactor(int n, int k) {
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int i = 1;
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for (; i < n / i; ++i) {
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if (n % i == 0 && (--k == 0)) {
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return i;
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}
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}
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if (i * i != n) {
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--i;
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}
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for (; i > 0; --i) {
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if (n % (n / i) == 0 && (--k == 0)) {
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return n / i;
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}
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}
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return -1;
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}
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};

‎solution/1400-1499/1492.The kth Factor of n/Solution.go

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func kthFactor(n int, k int) int {
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i := 1
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for ; i < n/i; i++ {
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if n%i == 0 {
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k--
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if k == 0 {
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return i
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}
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}
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}
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if i*i != n {
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i--
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}
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for ; i > 0; i-- {
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if n%(n/i) == 0 {
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k--
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if k == 0 {
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return n / i
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}
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}
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}
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return -1
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}

‎solution/1400-1499/1492.The kth Factor of n/Solution.java

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class Solution {
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public int kthFactor(int n, int k) {
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int i = 1;
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for (; i < n / i; ++i) {
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if (n % i == 0 && (--k == 0)) {
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return i;
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}
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}
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if (i * i != n) {
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--i;
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}
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for (; i > 0; --i) {
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if (n % (n / i) == 0 && (--k == 0)) {
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return n / i;
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}
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}
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return -1;
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}
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}

‎solution/1400-1499/1492.The kth Factor of n/Solution.py

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class Solution:
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def kthFactor(self, n: int, k: int) -> int:
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i = 1
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while i * i < n:
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if n % i == 0:
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k -= 1
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if k == 0:
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return i
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i += 1
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if i * i != n:
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i -= 1
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while i:
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if (n % (n // i)) == 0:
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k -= 1
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if k == 0:
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return n // i
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i -= 1
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return -1

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