(前缀和数组) lintcode 642. Moving Average from Data Stream

给定长度的sliding window, 求出数字流在这个window区间的平均值：

解法：前缀和数组：

index : 1 2 3 4

d : 1 10 3 5

sum: 1 11 14 19

sum[i] = d[1] + d[2] + ... + d[i]

d[x] + d[x+1] + ... + d[y] = sum[y] - sum[x-1]

优化：使用滚动数组来优化空间复杂度。

对于滑动窗口为3时，只存4个，即可以将下标变为 index %= 4

0 1 2 3 4 5 6 7 8 9

0 1 2 3 0 1 2 3 0 1

小技巧：滚动数组越长越不容易出错，所以取 >= size+1 就都可以

class MovingAverage {
public:
    /*
    * @param size: An integer
    */
    vector<double> sum;
    int id, size;
    MovingAverage(int size): sum(size+1, 0) {  //因为不知道sum数组的范围，故设一个大的值
        // do intialization if necessary
        id = 0;
        this->size = size;
    }
    
    int mod(int x){
        return x % (size+1);
    }
    /*
     * @param val: An integer
     * @return:  
     */
    double next(int val) {
        // write your code here
        id++;
        sum[mod(id)] = sum[mod(id-1)] + val;   //前缀和数组   //sum[0] = 0 为dummy node
        if(id - size >= 0){
            return (sum[mod(id)] - sum[mod(id-size)]) / size;
        }
        else
            return sum[mod(id)] / id;
    }
    
};

/**
 * Your MovingAverage object will be instantiated and called as such:
 * MovingAverage obj = new MovingAverage(size);
 * double param = obj.next(val);
 */

原文地址：https://www.cnblogs.com/Bella2017/p/11439840.html