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yanglbmeyanglbme
committedMar 16, 2023
feat: add solutions to lc problem: No.1362
No.1362.Closest Divisors
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‎solution/1300-1399/1362.Closest Divisors/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举**
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我们设计一个函数 $f(x)$，该函数返回乘积等于 $x$ 的两个数，且这两个数的差的绝对值最小。我们可以从 $\sqrt{x}$ 开始枚举 $i$，如果 $x$ 能被 $i$ 整除，那么 $\frac{x}{i}$ 就是另一个因数，此时我们就找到了一个乘积等于 $x$ 的两个因数，我们将其返回即可。否则我们减小 $i$ 的值，继续枚举。
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接下来，我们只需要分别计算 $f(num + 1)$ 和 $f(num + 2)$，然后比较两个函数的返回值，返回差的绝对值更小的那个即可。
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时间复杂度 $O(\sqrt{num})$，空间复杂度 $O(1)$。其中 $num$ 是给定的整数。
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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 closestDivisors(self, num: int) -> List[int]:
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def f(x):
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for i in range(int(sqrt(x)), 0, -1):
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if x % i == 0:
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return [i, x // i]
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a = f(num + 1)
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b = f(num + 2)
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return a if abs(a[0] - a[1]) < abs(b[0] - b[1]) else b
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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[] closestDivisors(int num) {
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int[] a = f(num + 1);
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int[] b = f(num + 2);
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return Math.abs(a[0] - a[1]) < Math.abs(b[0] - b[1]) ? a : b;
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}
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private int[] f(int x) {
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for (int i = (int) Math.sqrt(x);; --i) {
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if (x % i == 0) {
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return new int[] {i, x / i};
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}
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}
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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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vector<int> closestDivisors(int num) {
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auto f = [](int x) {
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for (int i = sqrt(x);; --i) {
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if (x % i == 0) {
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return vector<int>{i, x / i};
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}
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}
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};
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vector<int> a = f(num + 1);
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vector<int> b = f(num + 2);
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return abs(a[0] - a[1]) < abs(b[0] - b[1]) ? a : b;
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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 closestDivisors(num int) []int {
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f := func(x int) []int {
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for i := int(math.Sqrt(float64(x))); ; i-- {
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if x%i == 0 {
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return []int{i, x / i}
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}
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}
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}
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a, b := f(num+1), f(num+2)
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if abs(a[0]-a[1]) < abs(b[0]-b[1]) {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **...**

‎solution/1300-1399/1362.Closest Divisors/README_EN.md

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### **Python3**
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```python
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class Solution:
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def closestDivisors(self, num: int) -> List[int]:
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def f(x):
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for i in range(int(sqrt(x)), 0, -1):
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if x % i == 0:
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return [i, x // i]
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a = f(num + 1)
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b = f(num + 2)
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return a if abs(a[0] - a[1]) < abs(b[0] - b[1]) else b
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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[] closestDivisors(int num) {
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int[] a = f(num + 1);
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int[] b = f(num + 2);
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return Math.abs(a[0] - a[1]) < Math.abs(b[0] - b[1]) ? a : b;
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}
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private int[] f(int x) {
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for (int i = (int) Math.sqrt(x);; --i) {
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if (x % i == 0) {
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return new int[] {i, x / i};
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}
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}
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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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vector<int> closestDivisors(int num) {
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auto f = [](int x) {
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for (int i = sqrt(x);; --i) {
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if (x % i == 0) {
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return vector<int>{i, x / i};
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}
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}
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};
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vector<int> a = f(num + 1);
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vector<int> b = f(num + 2);
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return abs(a[0] - a[1]) < abs(b[0] - b[1]) ? a : b;
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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 closestDivisors(num int) []int {
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f := func(x int) []int {
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for i := int(math.Sqrt(float64(x))); ; i-- {
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if x%i == 0 {
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return []int{i, x / i}
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}
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}
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}
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a, b := f(num+1), f(num+2)
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if abs(a[0]-a[1]) < abs(b[0]-b[1]) {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **...**

‎solution/1300-1399/1362.Closest Divisors/Solution.cpp

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class Solution {
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public:
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vector<int> closestDivisors(int num) {
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auto f = [](int x) {
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for (int i = sqrt(x);; --i) {
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if (x % i == 0) {
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return vector<int>{i, x / i};
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}
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}
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};
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vector<int> a = f(num + 1);
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vector<int> b = f(num + 2);
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return abs(a[0] - a[1]) < abs(b[0] - b[1]) ? a : b;
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}
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};

‎solution/1300-1399/1362.Closest Divisors/Solution.go

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func closestDivisors(num int) []int {
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f := func(x int) []int {
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for i := int(math.Sqrt(float64(x))); ; i-- {
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if x%i == 0 {
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return []int{i, x / i}
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}
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}
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}
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a, b := f(num+1), f(num+2)
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if abs(a[0]-a[1]) < abs(b[0]-b[1]) {
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return a
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}
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return b
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}

‎solution/1300-1399/1362.Closest Divisors/Solution.java

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class Solution {
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public int[] closestDivisors(int num) {
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int[] a = f(num + 1);
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int[] b = f(num + 2);
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return Math.abs(a[0] - a[1]) < Math.abs(b[0] - b[1]) ? a : b;
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}
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private int[] f(int x) {
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for (int i = (int) Math.sqrt(x);; --i) {
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if (x % i == 0) {
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return new int[] {i, x / i};
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}
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}
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}
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}

‎solution/1300-1399/1362.Closest Divisors/Solution.py

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class Solution:
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def closestDivisors(self, num: int) -> List[int]:
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def f(x):
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for i in range(int(sqrt(x)), 0, -1):
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if x % i == 0:
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return [i, x // i]
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a = f(num + 1)
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b = f(num + 2)
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return a if abs(a[0] - a[1]) < abs(b[0] - b[1]) else b

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