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[6/1] 1480. Running Sum of 1d Array
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1480. Running Sum of 1d Array
Easy
2927203Add to ListShareGiven 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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링크
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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.
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 nums학부유학생님의 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
