Monday, May 2, 2016

LeetCode 346 - Moving Average from Data Stream


http://www.bubufx.com/detail-1431945.html
Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window.

For example,
MovingAverage m = new MovingAverage(3);
m.next(1) = 1
m.next(10) = (1 + 10) / 2
m.next(3) = (1 + 10 + 3) / 3
m.next(5) = (10 + 3 + 5) / 3
https://discuss.leetcode.com/topic/6/moving-average-from-data-stream/4
ublic class MovingAverage {
    private int size;
    private int sum;
    private Queue<Integer> window;

    public MovingAverage(int size) {
        this.size = size;
        this.sum = 0;
        this.window = new ArrayDeque<>();
    }

    public double next(int val) {
        window.offer(val);
        if (window.size() > size) {
            sum -= window.poll();
        }
        sum += val;
        return (double) sum / window.size();
    }
}
https://thorcsblog.wordpress.com/2016/05/01/leetcode-moving-average-from-data-stream/
public class MovingAverage {
    private int count;
    private int size;
    private int sum;
    private Queue<Integer> queue;
     
    /** Initialize your data structure here. */
    public MovingAverage(int size) {
        this.count = 0;
        this.size = size;
        this.sum = 0;
        this.queue = new LinkedList<>();
    }
     
    public double next(int val) {
        count++;
        if (count > size) {
            sum -= queue.poll();
        }
         
        queue.offer(val);
        sum += val;
         
        return count > size ? sum * 1.0 / size : sum * 1.0 / count;
    }
}
这道题定义了一个MovingAverage类,里面可以存固定个数字,然后我们每次读入一个数字,如果加上这个数字后总个数大于限制的个数,那么我们移除最早进入的数字,然后返回更新后的平均数,这种先进先出的特性最适合使用队列queue来做,而且我们还需要一个double型的变量sum来记录当前所有数字之和,这样有新数字进入后,如果没有超出限制个数,则sum加上这个数字,如果超出了,那么sum先减去最早的数字,再加上这个数字,然后返回sum除以queue的个数即可:
    MovingAverage(int size) {
        this->size = size;
        sum = 0;
    }
    
    double next(int val) {
        if (q.size() >= size) {
            sum -= q.front(); q.pop();
        }
        q.push(val);
        sum += val;
        return sum / q.size();
    }
Deque<Integer> dq; int size; int sum; public MovingAverage(int size) { dq = new LinkedList<>(); this.size = size; this.sum = 0; } public double next(int val) { if (dq.size() < size) { sum += val; dq.addLast(val); return (double) (sum / dq.size()); } else { int temp = dq.pollFirst(); sum -= temp; dq.addLast(val); sum += val; return (double) (sum / size); } }
X. 
private int [] window; private int n, insert; private long sum; /** Initialize your data structure here. */ public MovingAverage(int size) { window = new int[size]; insert = 0; sum = 0; } public double next(int val) { if (n < window.length) n++; sum -= window[insert]; sum += val; window[insert] = val; insert = (insert + 1) % window.length; return (double)sum / n; }

public class MovingAverage {
 
    LinkedList<Integer> queue;
    int size;
 
    /** Initialize your data structure here. */
    public MovingAverage(int size) {
        this.queue = new LinkedList<Integer>();
        this.size = size;
    }
 
    public double next(int val) {
        queue.offer(val);
        if(queue.size()>this.size){
            queue.poll();
        }
        int sum=0; // not efficient
        for(int i: queue){
            sum=sum+i;
        }
 
        return (double)sum/queue.size();
    }
}

Use ArrayDeque as Circular Array

    transient Object[] elements; // non-private to simplify nested class access
    public void addLast(E e) {
        if (e == null)
            throw new NullPointerException();
        elements[tail] = e;
        if ( (tail = (tail + 1) & (elements.length - 1)) == head)
            doubleCapacity();
    }
    public E pollLast() {
        int t = (tail - 1) & (elements.length - 1);
        @SuppressWarnings("unchecked")
        E result = (E) elements[t];
        if (result == null)
            return null;
        elements[t] = null;
        tail = t;
        return result;
    }

    public void addFirst(E e) {
        if (e == null)
            throw new NullPointerException();
        elements[head = (head - 1) & (elements.length - 1)] = e;
        if (head == tail)
            doubleCapacity();
    }

    public E pollFirst() {
        int h = head;
        @SuppressWarnings("unchecked")
        E result = (E) elements[h];
        // Element is null if deque empty
        if (result == null)
            return null;
        elements[h] = null;     // Must null out slot
        head = (h + 1) & (elements.length - 1);
        return result;
    }

    private void doubleCapacity() {
        assert head == tail;
        int p = head;
        int n = elements.length;
        int r = n - p; // number of elements to the right of p
        int newCapacity = n << 1;
        if (newCapacity < 0)
            throw new IllegalStateException("Sorry, deque too big");
        Object[] a = new Object[newCapacity];
        System.arraycopy(elements, p, a, 0, r);
        System.arraycopy(elements, 0, a, r, p);
        elements = a;
        head = 0;
        tail = n;
    }

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