8/11 每日一題(找出最長的嚴格遞增子序列)

 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: 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.

Example 2:

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

 

Constraints:

• 1 <= nums.length <= 104
• -109 <= nums[i] <= 109

class Solution:

    def findLengthOfLCIS(self, nums: List[int]) -> int:

        if len(nums)==1:

            return 1

        cur=1

        cur_max=1

        for i in range(1,len(nums)):

            if nums[i] > nums[i-1]:

                cur+=1

            else:

                #遇到非遞增

                cur=1

            cur_max=max(cur,cur_max)

        return cur_max

--------------------

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        l, r, curMax = 0, 1, 1
        while r < len(nums):
            if nums[r] > nums[r-1]:
                curMax = max(curMax, r-l+1)
            else:
                l = r
            r += 1
        return curMax