Maximum ascending subarray sum of leetcode simple problem

subject

I'll give you an array of positive integers nums , return nums In a Ascending The maximum possible elements of the subarray and .
A subarray is a sequence of consecutive numbers in an array .
Known subarray [numsl, numsl+1, …, numsr-1, numsr] , If for all i（l <= i < r）,numsi < numsi+1 All set up , This subarray is called Ascending Subarray . Be careful , The size is 1 The subarray of is also treated as Ascending Subarray .
Example 1：
Input ：nums = [10,20,30,5,10,50]
Output ：65
explain ：[5,10,50] Is the element and the largest ascending subarray , The largest sum of elements is 65 .
Example 2：
Input ：nums = [10,20,30,40,50]
Output ：150
explain ：[10,20,30,40,50] Is the element and the largest ascending subarray , The largest sum of elements is 150 .
Example 3：
Input ：nums = [12,17,15,13,10,11,12]
Output ：33
explain ：[10,11,12] Is the element and the largest ascending subarray , The largest sum of elements is 33 .
Example 4：
Input ：nums = [100,10,1]
Output ：100
Tips ：
1 <= nums.length <= 100
1 <= nums[i] <= 100
source ： Power button （LeetCode）

Their thinking

   There may be multiple ascending subarrays in an array , Traverse the array to check whether the elements are in ascending order and calculate the sum , If the element does not conform to ascending order , It means that the current ascending sub array has been traversed , Start traversing the next ascending subarray .

class Solution:
    def maxAscendingSum(self, nums: List[int]) -> int:
        s=nums[0]
        MAX=nums[0]
        for i in range(1,len(nums)):
            if nums[i]>nums[i-1]:
                s+=nums[i]
            else:
                if MAX<s:
                    MAX=s
                s=nums[i]
        return max(s,MAX)