Longest Continuous Increasing Subsequence

Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).

Example 1: Input: [1,3,5,4,7] Output: 3 Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. Even though [1,3,5,7] is also an increasing subsequence, it's not a continuous one where 5 and 7 are separated by 4.

Example 2: Input: [2,2,2,2,2] Output: 1 Explanation: The longest continuous increasing subsequence is [2], its length is 1.

Note: Length of the array will not exceed 10,000.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/longest-continuous-increasing-subsequence ## Basic Ideas: array has been splited as sections with increasing subsequence ## ## ## Complexity: Time O(n), Space O(1) class Solution: def findLengthOfLCIS(self, nums): """ :type nums: List[int] :rtype: int """ res, cnt = 0, 0 for i in range(0, len(nums)): if i==0 or nums[i-1] < nums[i]: cnt += 1 res = max(res, cnt) else: cnt = 1 return res # s = Solution() # print(s.findLengthOfLCIS([1,3,5,7])) # 4 # print(s.findLengthOfLCIS([1,3,5,4,7])) # 3 # print(s.findLengthOfLCIS([2,2,2,2])) # 1

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