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54 lines
1.1 KiB
JavaScript
54 lines
1.1 KiB
JavaScript
/*
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* 1480. Running Sum of 1d Array
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* Difficulty: Easy
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* https://leetcode.com/problems/running-sum-of-1d-array/
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*
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* ──────────────────────────────────────────────────
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*
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* Given an array nums. We define a running sum of an array as
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* runningSum[i] = sum(nums[0]…nums[i]).
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*
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* Return the running sum of nums.
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*
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*
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*
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* Example 1:
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*
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* Input: nums = [1,2,3,4]
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* Output: [1,3,6,10]
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* Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3,
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* 1+2+3+4].
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*
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* Example 2:
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*
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* Input: nums = [1,1,1,1,1]
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* Output: [1,2,3,4,5]
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* Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1,
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* 1+1+1+1, 1+1+1+1+1].
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*
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* Example 3:
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*
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* Input: nums = [3,1,2,10,1]
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* Output: [3,4,6,16,17]
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*
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*
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*
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* Constraints:
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*
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* • 1 <= nums.length <= 1000
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*
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* • -10^6 <= nums[i] <= 10^6
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*/
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/**
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* @param {number[]} nums
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* @return {number[]}
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*/
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var runningSum = function(nums) {
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let prefix = [nums[0]];
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for (let i = 1; i < nums.length; i++){
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prefix.push(prefix[i - 1] + nums[i]);
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}
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return prefix;
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};
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