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Commit 28073f8

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feat: add solutions to lc problem: No.0231 (#4631)
No.0231.Power of Two
1 parent 5e77929 commit 28073f8

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6 files changed

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‎solution/0200-0299/0231.Power of Two/README.md

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### 方法一:位运算
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$\texttt{n\&(n-1)}$ 可将最后一个二进制形式的 $n$ 的最后一位 1ドル$ 移除,若移除后为 0ドル,ドル说明 $n$ 是 2ドル$ 的幂。
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根据位运算的性质,执行 $\texttt{n\&(n-1)}$ 可以消去二进制形式的 $n$ 的最后一位 1ドル$。因此,如果 $n \gt 0,ドル并且满足 $\texttt{n\&(n-1)}$ 结果为 0ドル,ドル则说明 $n$ 是 2ドル$ 的幂。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn is_power_of_two(n: i32) -> bool {
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n > 0 && (n & (n - 1)) == 0
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}
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}
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```
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#### JavaScript
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```js
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<!-- solution:start -->
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### 方法二:lowbit
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### 方法二:Lowbit
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$\texttt{n\&(-n)}$ 可以得到 $n$ 的最后一位 1ドル$ 表示的十进制数,若与 $n$ 相等,说明 $n$ 是 2ドル$ 的幂。
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根据 $\text{lowbit}$ 的定义,我们知道 $\text{lowbit}(x) = x \& (-x),ドル可以得到 $n$ 的最后一位 1ドル$ 表示的十进制数。因此,如果 $n > 0,ドル并且满足 $\text{lowbit}(n)$ 等于 $n,ドル则说明 $n$ 是 2ドル$ 的幂。
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注意:要满足 $n$ 是 2ドル$ 的幂次方,需要保证 $n$ 大于 0ドル$。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```ts
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function isPowerOfTwo(n: number): boolean {
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return n > 0 && (n & (n - 1)) === 0;
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return n > 0 && n === (n & -n);
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn is_power_of_two(n: i32) -> bool {
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n > 0 && n == (n & (-n))
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}
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}
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```
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* @return {boolean}
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*/
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var isPowerOfTwo = function (n) {
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return n > 0 && n == (n & -n);
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return n > 0 && n === (n & -n);
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};
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```
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‎solution/0200-0299/0231.Power of Two/README_EN.md

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Bit Manipulation
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According to the properties of bit manipulation, executing $\texttt{n\&(n-1)}$ can eliminate the last bit 1ドル$ in the binary form of $n$. Therefore, if $n > 0$ and $\texttt{n\&(n-1)}$ results in 0ドル,ドル then $n$ is a power of 2ドル$.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn is_power_of_two(n: i32) -> bool {
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n > 0 && (n & (n - 1)) == 0
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}
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}
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```
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#### JavaScript
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```js
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<!-- solution:start -->
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### Solution 2
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### Solution 2: Lowbit
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According to the definition of $\text{lowbit},ドル we know that $\text{lowbit}(x) = x \& (-x),ドル which can get the decimal number represented by the last bit 1ドル$ of $n$. Therefore, if $n > 0$ and $\text{lowbit}(n)$ equals $n,ドル then $n$ is a power of 2ドル$.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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<!-- tabs:start -->
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```ts
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function isPowerOfTwo(n: number): boolean {
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return n > 0 && (n & (n - 1)) === 0;
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return n > 0 && n === (n & -n);
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn is_power_of_two(n: i32) -> bool {
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n > 0 && n == (n & (-n))
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}
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}
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```
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* @return {boolean}
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*/
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var isPowerOfTwo = function (n) {
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return n > 0 && n == (n & -n);
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return n > 0 && n === (n & -n);
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};
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```
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impl Solution {
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pub fn is_power_of_two(n: i32) -> bool {
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n > 0 && (n & (n - 1)) == 0
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}
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}

‎solution/0200-0299/0231.Power of Two/Solution2.js

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* @return {boolean}
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*/
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var isPowerOfTwo = function (n) {
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return n > 0 && n == (n & -n);
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return n > 0 && n === (n & -n);
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};
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impl Solution {
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pub fn is_power_of_two(n: i32) -> bool {
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n > 0 && n == (n & (-n))
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}
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}
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function isPowerOfTwo(n: number): boolean {
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return n > 0 && (n& (n -1))===0;
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return n > 0 && n=== (n &-n);
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}

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