Unlike choice sort, heapsort makes not waste time with a linear-time scan of the unsorted region; rather, heap sort keeps the unsorted region in a heap data structure to more quickly find the largest component in each step. Heapsort can be thought of as an better choice sort: like choice sort, heapsort divides its input into a sorted and an unsorted region, and it iteratively shrinks the unsorted region by extracting the largest component from it and inserting it into the sorted region.

The algorithm then repeatedly swaps the first value of the list with the last value, decrease the range of values considered in the heap operation by one, and sifting the new first value into its position in the heap. This repeats until the range of considered values is one value in length.

```
/// Sort a mutable slice using heap sort.
///
/// Heap sort is an in-place O(n log n) sorting algorithm. It is based on a
/// max heap, a binary tree data structure whose main feature is that
/// parent nodes are always greater or equal to their child nodes.
///
/// # Max Heap Implementation
///
/// A max heap can be efficiently implemented with an array.
/// For example, the binary tree:
/// ```text
/// 1
/// 2 3
/// 4 5 6 7
/// ```
///
/// ... is represented by the following array:
/// ```text
/// 1 23 4567
/// ```
///
/// Given the index `i` of a node, parent and child indices can be calculated
/// as follows:
/// ```text
/// parent(i) = (i-1) / 2
/// left_child(i) = 2*i + 1
/// right_child(i) = 2*i + 2
/// ```
/// # Algorithm
///
/// Heap sort has two steps:
/// 1. Convert the input array to a max heap.
/// 2. Partition the array into heap part and sorted part. Initially the
/// heap consists of the whole array and the sorted part is empty:
/// ```text
/// arr: [ heap |]
/// ```
///
/// Repeatedly swap the root (i.e. the largest) element of the heap with
/// the last element of the heap and increase the sorted part by one:
/// ```text
/// arr: [ root ... last | sorted ]
/// --> [ last ... | root sorted ]
/// ```
///
/// After each swap, fix the heap to make it a valid max heap again.
/// Once the heap is empty, `arr` is completely sorted.
pub fn heap_sort<T: Ord>(arr: &mut [T]) {
if arr.len() <= 1 {
return; // already sorted
}
heapify(arr);
for end in (1..arr.len()).rev() {
arr.swap(0, end);
move_down(&mut arr[..end], 0);
}
}
/// Convert `arr` into a max heap.
fn heapify<T: Ord>(arr: &mut [T]) {
let last_parent = (arr.len() - 2) / 2;
for i in (0..=last_parent).rev() {
move_down(arr, i);
}
}
/// Move the element at `root` down until `arr` is a max heap again.
///
/// This assumes that the subtrees under `root` are valid max heaps already.
fn move_down<T: Ord>(arr: &mut [T], mut root: usize) {
let last = arr.len() - 1;
loop {
let left = 2 * root + 1;
if left > last {
break;
}
let right = left + 1;
let max = if right <= last && arr[right] > arr[left] {
right
} else {
left
};
if arr[max] > arr[root] {
arr.swap(root, max);
}
root = max;
}
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn empty() {
let mut arr: Vec<i32> = Vec::new();
heap_sort(&mut arr);
assert_eq!(&arr, &[]);
}
#[test]
fn single_element() {
let mut arr = vec![1];
heap_sort(&mut arr);
assert_eq!(&arr, &[1]);
}
#[test]
fn sorted_array() {
let mut arr = vec![1, 2, 3, 4];
heap_sort(&mut arr);
assert_eq!(&arr, &[1, 2, 3, 4]);
}
#[test]
fn unsorted_array() {
let mut arr = vec![3, 4, 2, 1];
heap_sort(&mut arr);
assert_eq!(&arr, &[1, 2, 3, 4]);
}
#[test]
fn odd_number_of_elements() {
let mut arr = vec![3, 4, 2, 1, 7];
heap_sort(&mut arr);
assert_eq!(&arr, &[1, 2, 3, 4, 7]);
}
#[test]
fn repeated_elements() {
let mut arr = vec![542, 542, 542, 542];
heap_sort(&mut arr);
assert_eq!(&arr, &vec![542, 542, 542, 542]);
}
}
```