lintcode 1: Data Stream Median

Data Stream Median

Numbers keep coming, return the median of numbers at every time a new number added.

Have you met this question in a real interview?

Example

For numbers coming list: [1, 2, 3, 4, 5], return [1, 1, 2, 2, 3].

For numbers coming list: [4, 5, 1, 3, 2, 6, 0], return [4, 4, 4, 3, 3, 3, 3].

For numbers coming list: [2, 20, 100], return [2, 2, 20].

Challenge

Total run time in O(nlogn).

Clarification

What‘s the definition of Median?

- Median is the number that in the middle of a sorted array. If there are n numbers in a sorted array A, the median isA[(n - 1) / 2]. For example, if
A=[1,2,3], median is 2. If A=[1,19], median is
1.

[思路]

用两个堆, max heap 和 min heap. 维持两个堆的大小相等(max堆能够比min堆多一个).  则max堆的顶即为median值.

[CODE]

public class Solution {
    /**
     * @param nums: A list of integers.
     * @return: the median of numbers
     */
    public int[] medianII(int[] nums) {
        // write your code here
        if(nums==null) return null;
        int[] res = new int[nums.length];

        PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>();
        PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(11, new Comparator<Integer>() {
            @Override
            public int compare(Integer x, Integer y) {
                return y-x;
            }
        });
        res[0] = nums[0];
        maxHeap.add(nums[0]);

        for(int i=1; i<nums.length; i++) {
            int x = maxHeap.peek();
            if(nums[i] <= x) {
                maxHeap.add(nums[i]);
            } else {
                minHeap.add(nums[i]);
            }
            if(maxHeap.size() > minHeap.size()+1 ) {
                minHeap.add(maxHeap.poll());
            } else if(maxHeap.size() < minHeap.size()) {
                maxHeap.add(minHeap.poll());
            }
            res[i] = maxHeap.peek();
        }
        return res;
    }
}