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Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window.
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
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:
class MovingAverage { public: 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(); } private: queue<int> q; int size; double sum; };
类似题目:
https://leetcode.com/problems/moving-average-from-data-stream/
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered:
you can simply use a vector which wraps when there are more than "size" of elements, much more cache friendly.
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Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window.
Example:
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:
类似题目:
https://leetcode.com/problems/moving-average-from-data-stream/
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered: