力扣-295.数据流的中位数

题目链接

295.数据流的中位数

class MedianFinder {PriorityQueue<Integer> left;//队头最大PriorityQueue<Integer> right;//队头最小public MedianFinder() {left = new PriorityQueue<>(new Comparator<Integer>() {@Overridepublic int compare(Integer o1, Integer o2) {return o2 - o1;}});right = new PriorityQueue<>();}public void addNum(int num) {if (left.size() <= right.size()) {left.offer(num);} else {right.offer(num);}if (!left.isEmpty() && !right.isEmpty() && left.peek() > right.peek()) {left.offer(right.poll());right.offer(left.poll());}}public double findMedian() {if (left.size() > right.size()) {return left.peek();} else {return (left.peek() + right.peek()) / 2.0;}}
}

小结：两个优先级队列，左队列最大值不超过右队列最小值，且左队列容量=右队列(偶数)或左队列容量=右队列+1(奇数)，这样就可以根据两个队列的队头元素直接计算出中位数。