Problem Solving
-----------Problem-----------
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
------------Solution-----------
Solution: 01
class Solution {
public int[] runningSum(int[] nums) {
int[] output = new int[nums.length];
output[0] = nums[0];
for(int i = 1; i<nums.length; i++ ){
output[i]= nums[i] + output[i - 1] ;
System.out.println(output[i]);
}
return output;
}
}
Solution: 02
class Solution {
public int[] runningSum(int[] nums) {
for (int i= 1; i < nums.length; i++) {
nums[i] += nums[i - 1];
System.out.println(nums[i]);
};
return nums;
}
}
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