Problem statement

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

Solution

Just do what is asked: use two pointers beg and end, increase end if next number is greater than current; make both beg and end equal to end + 1 in opposite case.

Complexity

Time complexity is O(n), space complexity is O(1).

Code

class Solution:
    def findLengthOfLCIS(self, nums):
        beg, end, ans = 0, 0, int(len(nums) > 0)
        while end < len(nums) - 1:
            if nums[end] < nums[end+1]:
                end += 1
            else:
                beg, end = end + 1, end + 1
            ans = max(ans, end - beg + 1)
        return ans