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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
public int[] runningSum(int[] nums) {
int i = 1;
while (i<nums.length){
nums[i]+=nums[i-1];
i++;
}
return nums;
}
vector<int> runningSum(vector<int>& nums) {
int i = 1;
while (i<nums.size()){
nums[i]+=nums[i-1];
i++;
}
return nums;
}
def runningSum(self, nums):
i = 1
while i<len(nums):
nums[i]+=nums[i-1]
i+=1
return nums
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