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