feat: solve No.1800

BaffinLee · BaffinLee · commit 7b81795f5465 · 2023-05-14T22:26:47.000+08:00
diff --git a/1701-1800/1800. Maximum Ascending Subarray Sum.md b/1701-1800/1800. Maximum Ascending Subarray Sum.md
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+# 1800. Maximum Ascending Subarray Sum
+
+- Difficulty: Easy.
+- Related Topics: Array.
+- Similar Questions: Find Good Days to Rob the Bank, Maximum Number of Books You Can Take, Count Strictly Increasing Subarrays.
+
+## Problem
+
+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`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxAscendingSum = function(nums) {
+    var maxSum = nums[0];
+    var currSum = nums[0];
+    for (var i = 1; i < nums.length; i++) {
+        if (nums[i - 1] < nums[i]) {
+            currSum += nums[i];
+        } else {
+            maxSum = Math.max(maxSum, currSum);
+            currSum = nums[i];
+        }
+    }
+    return Math.max(maxSum, currSum);
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).
