LeetCode-in-All
300. Longest Increasing Subsequence
Medium
Given an integer array nums, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].
Example 1:
Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4
Example 3:
Input: nums = [7,7,7,7,7,7,7]
Output: 1
Constraints:
1 <= nums.length <= 2500-104 <= nums[i] <= 104
Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?
Solution
-spec length_of_lis(Nums :: [integer()]) -> integer().
length_of_lis(Nums) ->
Pred = fun(V, Index) -> { {Index, V}, Index + 1} end,
{IVals, _} = lists:mapfoldl(Pred, 0, Nums),
[H|T] = IVals,
Tid = ets:new(lis, []),
length_of_lis([], T, H, Tid, 0).
length_of_lis(Before, [], {_, Val}, Tid, Max) ->
max(lis(Before, Val, Tid) + 1, Max);
length_of_lis(Before, [Next|After], Current={Index, Val}, Tid, Max) ->
Lis = lis(Before, Val, Tid) + 1,
ets:insert(Tid, {Index, Lis}),
length_of_lis([Current|Before], After, Next, Tid, max(Max, Lis)).
lis([], _, _) -> 0;
lis(L, N, Tid) ->
Opts = lists:filter(fun({I, V}) -> V < N end, L),
Pred = fun({I, _}) -> ets:lookup_element(Tid, I, 2) end,
case Opts of
[] -> 0;
_ ->
lists:max(lists:map(Pred, Opts))
end.