Given an array of positive integers nums, return the maximum possible sum of an ascending subarray in nums.
A subarray is defined as a contiguous sequence of numbers in an array.
A subarray [numsl, numsl+1, ..., numsr-1, numsr] is ascending if for all i where l <= i < r, numsi < numsi+1. Note that a subarray of size l is ascending.
Example 1
Input: nums = [10,20,30,5,10,50]
Output: 65
Explanation: [5,10,50] is the ascending subarray with the maximum sum of 65.
Example 2
Input: nums = [10,20,30,40,50]
Output: 150
Explanation: [10,20,30,40,50] is the ascending subarray with the maximum sum of 150.
Example 3
Input: nums = [12,17,15,13,10,11,12]
Output: 33
Explanation: [10,11,12] is the ascending subarray with the maximum sum of 33.
Constraints
- 1 <= nums.length <= 100
- 1 <= nums[i] <= 100
Solution
题意理解：全是正整数的数组，求其最大升序子数组和，注意这里定义的升序不是非递减序，要求后一项严格大于前一项。
应熟练掌握的在线处理思想
int maxAscendingSum(int* nums, int numsSize){
int i, ThisSum, MaxSum, LastElem;
ThisSum = MaxSum = LastElem = 0;
for (i = 0; i < numsSize; i++) {
if (nums[i] > LastElem) ThisSum += nums[i];
else ThisSum = nums[i];
if (ThisSum > MaxSum) MaxSum = ThisSum;
LastElem = nums[i];
}
return MaxSum;
}
