Q1480 Running Sum of 1d Array-简单

Running Sum of 1d Array

Question

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.

Constraints:

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

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]

Approach 1:

开始的时候想建立sum变量来储存前i项和，但这样造成了空间的浪费，可以直接改写input列表，将其作为输出。

Note: 如果面试遇到这个问题，问清楚面试官，是否可以修改传来的 nums 数组！

约束条件限制了nums非空，因此不用判断这点；限制了元素大小，因此不考虑传入错误字符的情况

class Solution(object):
    def runningSum(self, nums):
        for i in range(1,len(nums)):
            nums[i]=nums[i-1]+nums[i]
        return nums


def main():
    nums = [1,1,1,1,1]
    solution=Solution().runningSum(nums)
    print(solution)


if __name__ == "__main__":
    main()

复杂度分析

• 时间复杂度O(n)
• 空间复杂度O(n)