346. 数据流中的移动平均值 🔒

题目描述

给定一个整数数据流和一个窗口大小，根据该滑动窗口的大小，计算其所有整数的移动平均值。

实现 MovingAverage 类：

MovingAverage(int size) 用窗口大小 size 初始化对象。
double next(int val) 计算并返回数据流中最后 size 个值的移动平均值。

示例：

输入：
["MovingAverage", "next", "next", "next", "next"]
[[3], [1], [10], [3], [5]]
输出：
[null, 1.0, 5.5, 4.66667, 6.0]

解释：
MovingAverage movingAverage = new MovingAverage(3);
movingAverage.next(1); // 返回 1.0 = 1 / 1
movingAverage.next(10); // 返回 5.5 = (1 + 10) / 2
movingAverage.next(3); // 返回 4.66667 = (1 + 10 + 3) / 3
movingAverage.next(5); // 返回 6.0 = (10 + 3 + 5) / 3

提示：

1 <= size <= 1000
-10⁵ <= val <= 10⁵
最多调用 next 方法 10⁴ 次

解法

方法一：循环数组

我们定义一个变量 \(\textit{s}\)，用于计算当前最后 \(\textit{size}\) 个元素的和，用一个变量 \(\textit{cnt}\) 记录当前元素的总数。另外，我们用一个长度为 \(\textit{size}\) 的数组 \(\textit{data}\) 记录每个位置的元素对应的值。

调用 \(\textit{next}\) 函数时，我们先计算出 \(\textit{val}\) 要存放的下标 \(i\)，然后我们更新元素和 \(s\)，并且将下标 \(i\) 处的值设置为 \(\textit{val}\)，同时将元素的个数加一。最后，我们返回 \(\frac{s}{\min(\textit{cnt}, \textit{size})}\) 的值即可。

时间复杂度 \(O(1)\)，空间复杂度 \(O(n)\)，其中 \(n\) 是题目给定的整数 \(\textit{size}\)。

Python3JavaC++GoTypeScript

class MovingAverage:

    def __init__(self, size: int):
        self.s = 0
        self.data = [0] * size
        self.cnt = 0

    def next(self, val: int) -> float:
        i = self.cnt % len(self.data)
        self.s += val - self.data[i]
        self.data[i] = val
        self.cnt += 1
        return self.s / min(self.cnt, len(self.data))


# Your MovingAverage object will be instantiated and called as such:
# obj = MovingAverage(size)
# param_1 = obj.next(val)

class MovingAverage {
    private int s;
    private int cnt;
    private int[] data;

    public MovingAverage(int size) {
        data = new int[size];
    }

    public double next(int val) {
        int i = cnt % data.length;
        s += val - data[i];
        data[i] = val;
        ++cnt;
        return s * 1.0 / Math.min(cnt, data.length);
    }
}

/**
 * Your MovingAverage object will be instantiated and called as such:
 * MovingAverage obj = new MovingAverage(size);
 * double param_1 = obj.next(val);
 */

class MovingAverage {
public:
    MovingAverage(int size) {
        data.resize(size);
    }

    double next(int val) {
        int i = cnt % data.size();
        s += val - data[i];
        data[i] = val;
        ++cnt;
        return s * 1.0 / min(cnt, (int) data.size());
    }

private:
    int s = 0;
    int cnt = 0;
    vector<int> data;
};

/**
 * Your MovingAverage object will be instantiated and called as such:
 * MovingAverage* obj = new MovingAverage(size);
 * double param_1 = obj->next(val);
 */

type MovingAverage struct {
    s    int
    cnt  int
    data []int
}

func Constructor(size int) MovingAverage {
    return MovingAverage{
        data: make([]int, size),
    }
}

func (this *MovingAverage) Next(val int) float64 {
    i := this.cnt % len(this.data)
    this.s += val - this.data[i]
    this.data[i] = val
    this.cnt++
    return float64(this.s) / float64(min(this.cnt, len(this.data)))
}

/**
 * Your MovingAverage object will be instantiated and called as such:
 * obj := Constructor(size);
 * param_1 := obj.Next(val);
 */

class MovingAverage {
    private s: number = 0;
    private cnt: number = 0;
    private data: number[];

    constructor(size: number) {
        this.data = Array(size).fill(0);
    }

    next(val: number): number {
        const i = this.cnt % this.data.length;
        this.s += val - this.data[i];
        this.data[i] = val;
        this.cnt++;
        return this.s / Math.min(this.cnt, this.data.length);
    }
}

/**
 * Your MovingAverage object will be instantiated and called as such:
 * var obj = new MovingAverage(size)
 * var param_1 = obj.next(val)
 */

方法二：队列

我们可以使用一个队列 \(\textit{q}\) 来存储最后 \(\textit{size}\) 个元素，同时用一个变量 \(\textit{s}\) 来记录这 \(\textit{size}\) 个元素的和。

在调用 \(\textit{next}\) 函数时，我们首先判断队列 \(\textit{q}\) 的长度是否等于 \(\textit{size}\)，如果等于 \(\textit{size}\)，则将队列 \(\textit{q}\) 的头部元素出队，并且更新 \(\textit{s}\) 的值。然后将 \(\textit{val}\) 入队，并且更新 \(\textit{s}\) 的值。最后返回 \(\frac{s}{\text{len}(q)}\) 的值即可。