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Commit 60d86dc

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feat: add solutions to lc problem: No.1387 (doocs#3875)
No.1387.Sort Integers by The Power Value
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‎solution/1300-1399/1387.Sort Integers by The Power Value/README.md‎

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### 方法一:自定义排序
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我们先定义一个函数 $f(x),ドル表示将数字 $x$ 变成 1ドル$ 所需要的步数,也即是数字 $x$ 的权重。
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我们先定义一个函数 $\textit{f}(x),ドル表示将数字 $x$ 变成 1ドル$ 所需要的步数,也即是数字 $x$ 的权重。
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然后我们将区间 $[lo, hi]$ 内的所有数字按照权重升序排序,如果权重相同,按照数字自身的数值升序排序。
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然后我们将区间 $[\textit{lo}, \textit{hi}]$ 内的所有数字按照权重升序排序,如果权重相同,按照数字自身的数值升序排序。
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最后返回排序后的第 $k$ 个数字。
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时间复杂度 $O(n \times \log n \times M),ドル空间复杂度 $O(n)$。其中 $n$ 是区间 $[lo, hi]$ 内的数字个数,而 $M$ 是 $f(x)$ 的最大值,本题中 $M$ 最大为 178ドル$。
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时间复杂度 $O(n \times \log n \times M),ドル空间复杂度 $O(n)$。其中 $n$ 是区间 $[\textit{lo}, \textit{hi}]$ 内的数字个数,而 $M$ 是 $f(x)$ 的最大值,本题中 $M$ 最大为 178ドル$。
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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 get_kth(lo: i32, hi: i32, k: i32) -> i32 {
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let f = |mut x: i32| -> i32 {
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let mut ans = 0;
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while x != 1 {
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if x % 2 == 0 {
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x /= 2;
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} else {
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x = 3 * x + 1;
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}
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ans += 1;
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}
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ans
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};
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let mut nums: Vec<i32> = (lo..=hi).collect();
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nums.sort_by(|&x, &y| f(x).cmp(&f(y)));
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nums[(k - 1) as usize]
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1300-1399/1387.Sort Integers by The Power Value/README_EN.md‎

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### Solution 1: Custom Sorting
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First, we define a function $f(x),ドル which represents the number of steps required to change the number $x$ to 1ドル,ドル i.e., the weight of the number $x$.
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First, we define a function $\textit{f}(x),ドル which represents the number of steps required to transform the number $x$ into 1ドル,ドル i.e., the power value of the number $x$.
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Then, we sort all the numbers in the interval $[lo, hi]$ in ascending order of weight. If the weights are the same, we sort them in ascending order of the numbers themselves.
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Then, we sort all the numbers in the interval $[\textit{lo}, \textit{hi}]$ in ascending order based on their power values. If the power values are the same, we sort them in ascending order based on the numbers themselves.
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Finally, we return the $k$-th number after sorting.
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Finally, we return the $k$-th number in the sorted list.
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The time complexity is $O(n \times \log n \times M),ドル and the space complexity is $O(n)$. Where $n$ is the number of numbers in the interval $[lo, hi],ドル and $M$ is the maximum value of $f(x)$. In this problem, the maximum value of $M$ is 178ドル$.
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The time complexity is $O(n \times \log n \times M),ドル and the space complexity is $O(n)$. Here, $n$ is the number of numbers in the interval $[\textit{lo}, \textit{hi}],ドル and $M$ is the maximum value of $f(x)$, which is at most 178ドル$ in this problem.
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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 get_kth(lo: i32, hi: i32, k: i32) -> i32 {
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let f = |mut x: i32| -> i32 {
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let mut ans = 0;
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while x != 1 {
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if x % 2 == 0 {
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x /= 2;
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} else {
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x = 3 * x + 1;
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}
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ans += 1;
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}
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ans
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};
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let mut nums: Vec<i32> = (lo..=hi).collect();
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nums.sort_by(|&x, &y| f(x).cmp(&f(y)));
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nums[(k - 1) as usize]
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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impl Solution {
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pub fn get_kth(lo: i32, hi: i32, k: i32) -> i32 {
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let f = |mut x: i32| -> i32 {
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let mut ans = 0;
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while x != 1 {
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if x % 2 == 0 {
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x /= 2;
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} else {
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x = 3 * x + 1;
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}
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ans += 1;
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}
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ans
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};
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let mut nums: Vec<i32> = (lo..=hi).collect();
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nums.sort_by(|&x, &y| f(x).cmp(&f(y)));
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nums[(k - 1) as usize]
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}
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}

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