Find Median from Data Stream

ID: 81; hard

LintCode 炼码

Solution 1 (Java)

public class Solution {

    PriorityQueue<Integer> minHeap, maxHeap;
    int median;
    boolean isFirstNum;
    public Solution() {
        minHeap = new PriorityQueue<>();
        maxHeap = new PriorityQueue<>(new Comparator<Integer>() {
            public int compare(Integer x, Integer y) {
                return y - x;
            }
        });
        isFirstNum = true;
    }

    /**
     * @param val: An integer
     * @return: nothing
     */
    public void add(int val) {
        if (isFirstNum) {
            median = val;
            isFirstNum = false;
            return;
        }

        if (val < median) {
            maxHeap.offer(val);
        } else {
            minHeap.offer(val);
        }

        if (maxHeap.size() > minHeap.size()) {
            minHeap.offer(median);
            median = maxHeap.poll();
        }
        if (maxHeap.size() < minHeap.size() - 1) {
            maxHeap.offer(median);
            median = minHeap.poll();
        }
    }

    /**
     * @return: return the median of the data stream
     */
    public int getMedian() {
        return median;
    }
}

Notes

• We used a maxHeap and a minHeap to maintain the data stream. The maxHeap stores numbers that are less than the median. The minHeap stores numbers that are larger than the median. Then, we update the median whenever the two heaps have sizes that differ by 2 (to be more precise, see the condition in code). By construction, the results by polling either the minHeap or the maxHeap will be the current median. Eventually, we just return the current median.

• The reason is that we need to find the median of k numbers, so it is natural to find the k/2th largest number and the k/2th smallest number (first number in each heap).