文章目錄
- 1480.一維數組的動态和
-
- 題目描述
- 示例1
- 示例2
- 示例3
- 提示
- 思路:原地修改、整活
-
- 原地修改
- 整活
1480.一維數組的動态和
題目描述
給你一個數組nums。數組動态和的計算公式為:runningSum[i] = sum(nums[0]…nums[i])
請傳回nums的動态和。
示例1
輸入:nums = [1, 2, 3, 4]
輸出:[1, 3, 6, 10]
示例2
輸入:nums = [1, 1, 1, 1, 1]
輸出:[1, 2, 3, 4, 5]
示例3
輸入:nums = [3, 1, 2, 10, 1]
輸出:[3, 4, 6, 16, 17]
提示
- 1 <= nums.length <= 1000
- -106 <= nums[i] <= 106
思路:原地修改、整活
原地修改
![](https://img.laitimes.com/img/_0nNw4CM6IyYiwiM6ICdiwiIyVGduV2YfNWawNyROBlLwE2MmRzMiBTMxkTN4EmZ5czN5QzM4EDZ0UDN4ATNlNzLc52YucWbp5GZzNmLn9Gbi1yZtl2Lc9CX6MHc0RHaiojIsJye.png)
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
for i in range(1, len(nums)):
nums[i] += nums[i-1]
return nums
整活
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
return list(accumulate(nums))