【朱迪的LeetCode刷题笔记】1480. Running Sum of 1d Array #Easy #Python

Judie

Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).
Return the running sum of nums.

Example 1:
Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].

Example 2:
Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].

Example 3:
Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]


Constraints:
1 <= nums.length <= 1000
-10^6 <= nums[i] <= 10^6

1. class Solution(object):
   
2.     def runningSum(self, nums):
   
3.         """
   
4.         :type nums: List[int]
   
5.         :rtype: List[int]
   
6.         """

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Judy

1. class Solution(object):
   
2.     def runningSum(self, nums):
   
3.         """
   
4.         :type nums: List[int]
   
5.         :rtype: List[int]
   
6.         """
   
7.         i = 1
   
8.         l = len(nums)
   
9.         s = nums[0]
   
10.         runningSum = [s]
    
11.         while i < l:
    
12.             s += nums[i]
    
13.             runningSum.append(s)
    
14.             i +=1
    
15.         return runningSum
    
16.         

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Gray

1. class Solution(object):
   
2.     def runningSum(self, nums):
   
3.         """
   
4.         :type nums: List[int]
   
5.         :rtype: List[int]
   
6.         """
   
7.         ans = []
   
8.         temp = 0
   
9.         for x in nums:
   
10.             ans.append(temp + x)
    
11.             temp = temp + x
    
12.         return ans

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不二如是

不错，希望更下去

Judie

不二如是 发表于 2023-5-23 06:35
不错，希望更下去

谢谢 我努力！