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[6/1] 1480. Running Sum of 1d Array

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1480. Running Sum of 1d Array
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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

관련자료

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시엉준님의 댓글

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JAVA
Runtime: 2 ms, faster than 80.36% of Java online submissions for Valid Parentheses.
Memory Usage: 42.9 MB, less than 5.64% of Java online submissions for Valid Parentheses.

class Solution {
    public int[] runningSum(int[] nums) {
        for(int i=1; i<nums.length; i++) {
            nums[i] += nums[i-1];
        }
        
        return nums;
    }
}

나무토끼님의 댓글

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

학부유학생님의 댓글

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class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        total = 0
        res = []
        for num in nums:
            total += num
            res.append(total)
        
        return res
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