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feat: add solutions to lc problem: No.0932
No.0932.Beautiful Array
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‎solution/0900-0999/0932.Beautiful Array/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：分治**
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根据题意，漂亮数组 $A$ 需要满足对于任意 $i<k<j$, $A_k*2 \neq A_i+A_j$。
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我们可以发现，不等式左侧一定是偶数，那么我们只要保证不等式右侧 $A_i$ 和 $A_j$ 分别是一奇一偶，那么不等式就恒成立。
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利用分治，我们将 $n$ 缩小规模为原来的一半，递归调用，可以得到两个漂亮数组 $left$, $right$。我们将 $left$ 中每个元素 $x_i$ 变为 $x_i*2-1$ 可以得到一个奇数数组；将 $right$ 中每个元素 $x_i$ 变为 $x_i*2$，可以得到一个偶数数组。这两个数组仍然是漂亮数组。
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> 基于一个性质，将漂亮数组中的每个元素 $x_i$ 变换为 $kx_i+b$，得到的数组仍然是漂亮数组。
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将这两个漂亮数组合并在一起，由于满足一奇一偶，那么合并后的数组也是漂亮数组，从而得到了答案。
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时间复杂度 $O(nlogn)$。
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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 beautifulArray(self, n: int) -> List[int]:
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if n == 1:
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return [1]
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left = self.beautifulArray((n + 1) >> 1)
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right = self.beautifulArray(n >> 1)
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left = [x * 2 - 1 for x in left]
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right = [x * 2 for x in right]
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return left + right
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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[] beautifulArray(int n) {
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if (n == 1) {
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return new int[]{1};
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}
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int[] left = beautifulArray((n + 1) >> 1);
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int[] right = beautifulArray(n >> 1);
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int[] ans = new int[n];
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int i = 0;
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for (int x : left) {
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ans[i++] = x * 2 - 1;
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}
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for (int x : right) {
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ans[i++] = x * 2;
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}
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return ans;
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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> beautifulArray(int n) {
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if (n == 1) return {1};
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vector<int> left = beautifulArray((n + 1) >> 1);
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vector<int> right = beautifulArray(n >> 1);
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vector<int> ans(n);
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int i = 0;
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for (int& x : left) ans[i++] = x * 2 - 1;
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for (int& x : right) ans[i++] = x * 2;
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return ans;
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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 beautifulArray(n int) []int {
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if n == 1 {
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return []int{1}
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}
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left := beautifulArray((n + 1) >> 1)
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right := beautifulArray(n >> 1)
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var ans []int
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for _, x := range left {
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ans = append(ans, x*2-1)
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}
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for _, x := range right {
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ans = append(ans, x*2)
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}
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return ans
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}
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```
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### **...**

‎solution/0900-0999/0932.Beautiful Array/README_EN.md

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### **Python3**
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```python
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class Solution:
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def beautifulArray(self, n: int) -> List[int]:
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if n == 1:
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return [1]
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left = self.beautifulArray((n + 1) >> 1)
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right = self.beautifulArray(n >> 1)
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left = [x * 2 - 1 for x in left]
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right = [x * 2 for x in right]
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return left + right
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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[] beautifulArray(int n) {
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if (n == 1) {
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return new int[]{1};
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}
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int[] left = beautifulArray((n + 1) >> 1);
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int[] right = beautifulArray(n >> 1);
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int[] ans = new int[n];
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int i = 0;
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for (int x : left) {
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ans[i++] = x * 2 - 1;
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}
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for (int x : right) {
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ans[i++] = x * 2;
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}
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return ans;
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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> beautifulArray(int n) {
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if (n == 1) return {1};
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vector<int> left = beautifulArray((n + 1) >> 1);
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vector<int> right = beautifulArray(n >> 1);
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vector<int> ans(n);
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int i = 0;
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for (int& x : left) ans[i++] = x * 2 - 1;
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for (int& x : right) ans[i++] = x * 2;
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return ans;
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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 beautifulArray(n int) []int {
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if n == 1 {
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return []int{1}
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}
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left := beautifulArray((n + 1) >> 1)
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right := beautifulArray(n >> 1)
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var ans []int
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for _, x := range left {
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ans = append(ans, x*2-1)
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}
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for _, x := range right {
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ans = append(ans, x*2)
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}
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return ans
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}
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```
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### **...**

‎solution/0900-0999/0932.Beautiful Array/Solution.cpp

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class Solution {
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public:
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vector<int> beautifulArray(int n) {
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if (n == 1) return {1};
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vector<int> left = beautifulArray((n + 1) >> 1);
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vector<int> right = beautifulArray(n >> 1);
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vector<int> ans(n);
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int i = 0;
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for (int& x : left) ans[i++] = x * 2 - 1;
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for (int& x : right) ans[i++] = x * 2;
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return ans;
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}
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};

‎solution/0900-0999/0932.Beautiful Array/Solution.go

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func beautifulArray(n int) []int {
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if n == 1 {
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return []int{1}
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}
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left := beautifulArray((n + 1) >> 1)
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right := beautifulArray(n >> 1)
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var ans []int
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for _, x := range left {
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ans = append(ans, x*2-1)
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}
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for _, x := range right {
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ans = append(ans, x*2)
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}
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return ans
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}

‎solution/0900-0999/0932.Beautiful Array/Solution.java

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class Solution {
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public int[] beautifulArray(int n) {
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if (n == 1) {
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return new int[]{1};
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}
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int[] left = beautifulArray((n + 1) >> 1);
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int[] right = beautifulArray(n >> 1);
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int[] ans = new int[n];
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int i = 0;
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for (int x : left) {
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ans[i++] = x * 2 - 1;
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}
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for (int x : right) {
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ans[i++] = x * 2;
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}
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return ans;
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}
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}

‎solution/0900-0999/0932.Beautiful Array/Solution.py

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class Solution:
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def beautifulArray(self, n: int) -> List[int]:
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if n == 1:
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return [1]
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left = self.beautifulArray((n + 1) >> 1)
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right = self.beautifulArray(n >> 1)
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left = [x * 2 - 1 for x in left]
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right = [x * 2 for x in right]
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return left + right

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