<p>Given an unsorted array of integers <code>nums</code>, return <em>the length of the longest <strong>continuous increasing subsequence</strong> (i.e. subarray)</em>. The subsequence must be <strong>strictly</strong> increasing.</p>

<p>A <strong>continuous increasing subsequence</strong> is defined by two indices <code>l</code> and <code>r</code> (<code>l &lt; r</code>) such that it is <code>[nums[l], nums[l + 1], ..., nums[r - 1], nums[r]]</code> and for each <code>l &lt;= i &lt; r</code>, <code>nums[i] &lt; nums[i + 1]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,5,4,7]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        maxLength = 1
        currentLength = 1
        for i in range(len(nums)-1):
            if nums[i] < nums[i+1]:
                currentLength += 1
            else:
                currentLength = 1
            if currentLength > maxLength:
                maxLength = currentLength
        
        return maxLength
17900