【leetcode】1480. Running Sum of 1d Array

题目如下：

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]

解题思路：送分题。

代码如下：

class Solution(object):
    def runningSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        res = []
        count = 0
        for i in nums:
            count += i
            res.append(count)
        return res

【leetcode】1480. Running Sum of 1d Array

原文：https://www.cnblogs.com/seyjs/p/13176600.html