Window Sum

Given an array of n integer, and a moving window(size k), move the window at each iteration from the start of the array, find the sum of the element inside the window at each moving. 

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Example

For array [1,2,7,8,5], moving window size k = 3. 
1 + 2 + 7 = 10
2 + 7 + 8 = 17
7 + 8 + 5 = 20
return [10,17,20]

java

public class Solution {
    /*
     * @param nums: a list of integers.
     * @param k: length of window.
     * @return: the sum of the element inside the window at each moving.
     */
    public int[] winSum(int[] nums, int k) {
        // write your code here
        if (nums == null || nums.length == 0)  {
            return nums;
        }
        if (nums.length < k) {
            return null;
        }
        int[] temp = new int[nums.length + 1];
        temp[0] = 0;
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            temp[i + 1] = sum; 
        }
        int[] result = new int[nums.length - k + 1];
        for (int i = k; i < temp.length; i++) {
            result[i - k] = temp[i] - temp[i - k];
        }
        return result;
    }
}

python

class Solution:
    """
    @param: nums: a list of integers.
    @param: k: length of window.
    @return: the sum of the element inside the window at each moving.
    """
    def winSum(self, nums, k):
        # write your code here
        if nums is None or len(nums) == 0 or k <= 0:
            return nums
        if len(nums) < k:
            return None
        result = [0] * (len(nums) - k + 1)
        for i in range(k):
            result[0] += nums[i]
        for i in range(1, len(nums) - k + 1):
            result[i] = result[i - 1] - nums[i - 1] + nums[i + k - 1]
        return result