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Rollup merge of rust-lang#60952 - dtolnay:heap, r=Amanieu
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Document BinaryHeap time complexity

I went into some detail on the time complexity of `push` because it is relevant for using BinaryHeap efficiently -- specifically that you should avoid pushing many elements in ascending order when possible.

r? @Amanieu
Closes rust-lang#47976. Closes rust-lang#59698.
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Centril authored May 20, 2019
2 parents 65cec43 + ea7aa76 commit 864e7a9
Showing 1 changed file with 43 additions and 0 deletions.
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

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