lis
pub fn longest_increasing_subsequence<T: PartialOrd>(a: &[T]) -> Vec<usize>
Returns the indices of the longest increasing subsequence in a.
a
This is O(n log n).
Panics if a.len() is zero.
a.len()
use lis::longest_increasing_subsequence; assert_eq!(longest_increasing_subsequence(&[2, 1, 4, 3, 5]), [1, 3, 4]);