【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

class Solution(object):
    def runningSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """

Judy

class Solution(object):
    def runningSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        i = 1
        l = len(nums)
        s = nums[0]
        runningSum = [s]
        while i < l:
            s += nums[i]
            runningSum.append(s)
            i +=1
        return runningSum
        

Gray

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

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！