Problem:2090. 半径为 k 的子数组平均值

思路：定长滑动窗口或者前缀和

Accode:

class Solution {
public:
    vector<int> getAverages(vector<int>& nums, int k) {
        int len = nums.size();
        vector<int> ans(len,-1);
        long long sum = 0;
        for(int i=0;i<len;i++){
            if(i+1>2*k+1) sum-=nums[i-k-k-1];
            sum+=nums[i];
            if(i+1>=2*k+1) ans[i-k] = sum/(2*k+1);
        }
        return ans;
    }
};

时间复杂度:$o(n)$