## Description

https://leetcode.com/problems/longest-continuous-increasing-subsequence/

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`

and `r`

(`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]`

.

**Example 1:**

Input:nums = [1,3,5,4,7]Output:3Explanation: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.

**Example 2:**

Input:nums = [2,2,2,2,2]Output:1Explanation:The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly increasing.

**Constraints:**

`0 <= nums.length <= 10`

^{4}`-10`

^{9}<= nums[i] <= 10^{9}

## Explanation

Iterate the list and compare it with the previous number and reach a conclusion what is the longest continuous sequence.

## Python Solution

```
class Solution:
def findLengthOfLCIS(self, nums: List[int]) -> int:
max_count = 0
if not nums:
return max_count
prev = nums[0]
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).