QUESTION:
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 1 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
EXPLANATION:
只要判断出当前位置和前一个位置的大小, 如果比前一个位置大, 那么就在tmpresult上加上结果. 如果没有前一个位置大, 就将当前值传给tmpresult. 然后进行result和tmpresult的大小比较即可.
SOLUTION:
class Solution {
func maxAscendingSum(_ nums: [Int]) -> Int {
var result = 0
var tmpResult = 0
for index in nums.indices {
if (index == 0) {
tmpResult = nums[index]
} else {
if (nums[index] > nums[index - 1]) {
tmpResult += nums[index]
} else {
tmpResult = nums[index]
}
}
result = max(result, tmpResult)
}
return result
}
}