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Add partition_point #73577
Add partition_point #73577
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Original file line number | Diff line number | Diff line change |
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@@ -2663,6 +2663,60 @@ impl<T> [T] { | |
{ | ||
self.iter().is_sorted_by_key(f) | ||
} | ||
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/// Returns the index of the partition point according to the given predicate | ||
/// (the index of the first element of the second partition). | ||
/// | ||
/// The slice is assumed to be partitioned according to the given predicate. | ||
/// This means that all elements for which the predicate returns true are at the start of the slice | ||
/// and all elements for which the predicate returns false are at the end. | ||
/// For example, [7, 15, 3, 5, 4, 12, 6] is a partitioned under the predicate x % 2 != 0 | ||
/// (all odd numbers are at the start, all even at the end). | ||
/// | ||
/// If this slice is not partitioned, the returned result is unspecified and meaningless, | ||
/// as this method performs a kind of binary search. | ||
/// | ||
/// # Examples | ||
/// | ||
/// ``` | ||
/// #![feature(partition_point)] | ||
/// | ||
/// let v = [1, 2, 3, 3, 5, 6, 7]; | ||
/// let i = v.partition_point(|&x| x < 5); | ||
/// | ||
/// assert_eq!(i, 4); | ||
/// assert!(v[..i].iter().all(|&x| x < 5)); | ||
/// assert!(v[i..].iter().all(|&x| !(x < 5))); | ||
/// ``` | ||
#[unstable(feature = "partition_point", reason = "new API", issue = "73831")] | ||
pub fn partition_point<P>(&self, mut pred: P) -> usize | ||
where | ||
P: FnMut(&T) -> bool, | ||
{ | ||
let mut left = 0; | ||
let mut right = self.len(); | ||
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while left != right { | ||
let mid = left + (right - left) / 2; | ||
// SAFETY: | ||
// When left < right, left <= mid < right. | ||
// Therefore left always increases and right always decreases, | ||
// and eigher of them is selected. | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. I think There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. oops thank you. |
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// In both cases left <= right is satisfied. | ||
// Therefore if left < right in a step, | ||
// left <= right is satisfied in the next step. | ||
// Therefore as long as left != right, 0 <= left < right <= len is satisfied | ||
// and if this case 0 <= mid < len is satisfied too. | ||
let value = unsafe { self.get_unchecked(mid) }; | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Please add a comment about why this |
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if pred(value) { | ||
left = mid + 1; | ||
} else { | ||
right = mid; | ||
} | ||
} | ||
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left | ||
} | ||
} | ||
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#[lang = "slice_u8"] | ||
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I think we'd better change to "If this slice is not sorted ..." as we actually want to tell the end-user an ordered slice is required for this method.
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The comment is right -- the slice has to be partitioned, and not ordered. Raison d'être of
partition_point
is that it works with non-Ord
things. To make this more precise, we might link to https://doc.rust-lang.org/stable/std/iter/trait.Iterator.html#method.is_partitioned.There was a problem hiding this comment.
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Indeed in 95% cases
partition_point
is used to getlower/upper_bound
. However in 5% cases it looks more partition point than lower bound.For example, we can get all valid values from
[3, 1, 4, 1, 5, -1, -1, -1, -1, -1]
.Actually this array is sorted by
|&x| x != -1
in descending order (true is larger than false), but many user would think that the array is partitioned at index 5 by whether -1 or not.What do you think about adding like this?
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Thanks for the kind explanation and sorry for my misunderstanding. Yes, the slice requires partitioned instead of sorted.
However, I still confused by the comment
If this slice is not partitioned, the returned result is unspecified and meaningless
. If we did call this method on a not partitioned slice, the result should be anOption<size>
rather than an unspecified or meaningless value.There was a problem hiding this comment.
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Binary search runs in O(log n) by assuming ordered. If it checks whether is sorted, it must slower than O(n).