[daily 3 questions (2)] maximum ascending subarray sum

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

Ideas ：

One traverse , Record the cumulative sum of ascending order and the maximum value of accumulation , If found no longer ascending , Then recalculate the cumulative sum .

java Code ：

class Solution {
    public int maxAscendingSum(int[] nums) {
        int sum = nums[0];
        int max = nums[0];
        for (int i = 1; i < nums.length; i++) {
            if(nums[i]>nums[i-1]) {
                sum+=nums[i];
            }else {
                sum = nums[i];
            }
            max = Math.max(max, sum);
        }
        return max;
    }
}