題目：
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.

思路：

1. 先算第一個框框總和
2. 接下來只要把框框+後一個，-前一個，就可以得到一個新的總和

代碼：

public:
    /**
     * @param nums: a list of integers.
     * @param k: length of window.
     * @return: the sum of the element inside the window at each moving.
     */
    vector<int> winSum(vector<int> &nums, int k) {
        // write your code here
        if(nums.size()==0 || nums.size() < k || k == 0)
            return {};
        
        int currSum =0;
        vector<int> ans;
        
        for(int i =0; i<k; i++){
            currSum+= nums[i];
        }
        ans.push_back(currSum);
        
        for(int i=k; i<nums.size(); i++){
            currSum+= nums[i];
            currSum-= nums[i-k];
            ans.push_back(currSum);
        }
       
        return ans;
       
    }
};```