Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Document BinaryHeap time complexity #60952

Merged
merged 1 commit into from
May 21, 2019
Merged
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
43 changes: 43 additions & 0 deletions src/liballoc/collections/binary_heap.rs
Original file line number Diff line number Diff line change
Expand Up @@ -231,6 +231,20 @@ use super::SpecExtend;
/// assert_eq!(heap.pop(), Some(Reverse(5)));
/// assert_eq!(heap.pop(), None);
/// ```
///
/// # Time complexity
///
/// | [push] | [pop] | [peek]/[peek\_mut] |
/// |--------|----------|--------------------|
/// | O(1)~ | O(log n) | O(1) |
///
/// The value for `push` is an expected cost; the method documentation gives a
/// more detailed analysis.
///
/// [push]: #method.push
/// [pop]: #method.pop
/// [peek]: #method.peek
/// [peek\_mut]: #method.peek_mut
#[stable(feature = "rust1", since = "1.0.0")]
pub struct BinaryHeap<T> {
data: Vec<T>,
Expand Down Expand Up @@ -384,6 +398,10 @@ impl<T: Ord> BinaryHeap<T> {
/// }
/// assert_eq!(heap.peek(), Some(&2));
/// ```
///
/// # Time complexity
///
/// Cost is O(1) in the worst case.
#[stable(feature = "binary_heap_peek_mut", since = "1.12.0")]
pub fn peek_mut(&mut self) -> Option<PeekMut<'_, T>> {
if self.is_empty() {
Expand Down Expand Up @@ -411,6 +429,11 @@ impl<T: Ord> BinaryHeap<T> {
/// assert_eq!(heap.pop(), Some(1));
/// assert_eq!(heap.pop(), None);
/// ```
///
/// # Time complexity
///
/// The worst case cost of `pop` on a heap containing *n* elements is O(log
/// n).
#[stable(feature = "rust1", since = "1.0.0")]
pub fn pop(&mut self) -> Option<T> {
self.data.pop().map(|mut item| {
Expand Down Expand Up @@ -438,6 +461,22 @@ impl<T: Ord> BinaryHeap<T> {
/// assert_eq!(heap.len(), 3);
/// assert_eq!(heap.peek(), Some(&5));
/// ```
///
/// # Time complexity
///
/// The expected cost of `push`, averaged over every possible ordering of
/// the elements being pushed, and over a sufficiently large number of
/// pushes, is O(1). This is the most meaningful cost metric when pushing
/// elements that are *not* already in any sorted pattern.
///
/// The time complexity degrades if elements are pushed in predominantly
/// ascending order. In the worst case, elements are pushed in ascending
/// sorted order and the amortized cost per push is O(log n) against a heap
/// containing *n* elements.
///
/// The worst case cost of a *single* call to `push` is O(n). The worst case
/// occurs when capacity is exhausted and needs a resize. The resize cost
/// has been amortized in the previous figures.
#[stable(feature = "rust1", since = "1.0.0")]
pub fn push(&mut self, item: T) {
let old_len = self.len();
Expand Down Expand Up @@ -650,6 +689,10 @@ impl<T> BinaryHeap<T> {
/// assert_eq!(heap.peek(), Some(&5));
///
/// ```
///
/// # Time complexity
///
/// Cost is O(1) in the worst case.
#[stable(feature = "rust1", since = "1.0.0")]
pub fn peek(&self) -> Option<&T> {
self.data.get(0)
Expand Down