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描述
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 [nums[l], nums[l+1], ..., nums[r-1], nums[r]] is ascending if for all i where l <= i < r, nums[i] < nums[i+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.
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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.
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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.
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Example 4:
Input: nums = [100,10,1]
Output: 100
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Note:
1 <= nums.length <= 100
1 <= nums[i] <= 100
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解析
根据题意,就是给出一个正整数的列表 nums ,找出升序子列表的最大和。子序列就是列表 nums 中的若干个连续的元素组成的列表,升序列就是后一个元素的值比前一个元素的值要大,注意只有一个元素的子序列也算是升序列。思路:
- 如果 nums 长度为 1 ,nums 本身就是升序列,直接返回 nums[0]
- 否则使用 total 表示当前子列表的和, max_result 表示最终需要返回的最大和。遍历 nums ,如果当前的元素小于等于前一个元素,那么直接将 total 设置为当前元素,表示从这里开始往后找新的升序子列表。如果当前元素大于前一个元素,那么当前的升序列表还可以找下去,将该元素加入 total
- 然后比较 total 和 max_result 的最大值来更新 max_result
- 遍历结束即可得到最终的结果 max_result 。
解答
class Solution(object):
def maxAscendingSum(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 1:
return nums[0]
total = nums[0]
max_result = nums[0]
for i in range(1, len(nums)):
e = nums[i]
if e>nums[i-1]:
total += e
max_result = max(max_result, total)
else:
total = e
return max_result
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运行结果
Runtime: 16 ms, faster than 93.91% of Python online submissions for Maximum Ascending Subarray Sum.
Memory Usage: 13.6 MB, less than 9.68% of Python online submissions for Maximum Ascending Subarray Sum.
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原题链接:leetcode.com/problems/ma…
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