Given an unsorted array of integers
nums, return the length of the longest continuous increasing subsequence (i.e. subarray). The subsequence must be strictly increasing.
A continuous increasing subsequence is defined by two indices
l < r) such that it is
[nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] and for each
l <= i < r,
nums[i] < nums[i + 1].
Input: nums = [1,3,5,4,7] Output: 3 Explanation: The longest continuous increasing subsequence is [1,3,5] with length 3. Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element 4.
Input: nums = [2,2,2,2,2] Output: 1 Explanation: The longest continuous increasing subsequence is  with length 1. Note that it must be strictly increasing.
0 <= nums.length <= 104
-109 <= nums[i] <= 109
Iterate the list and compare it with the previous number and reach a conclusion what is the longest continuous sequence.
class Solution: def findLengthOfLCIS(self, nums: List[int]) -> int: max_count = 0 if not nums: return max_count prev = nums count = 1 for num in nums[1:]: if num > prev: count += 1 else: max_count = max(max_count, count) count = 1 prev = num return max(max_count, count)
- Time Complexity: O(N).
- Space Complexity: O(1).