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

Have you met this question in a real interview?

Yes

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, ifA=[1,2,3], median is2. IfA=[1,19], median is1.

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].

思路

暴力解法可以用二分查找插入保持数组的有序性，然后通过return中间的元素返回中位数， 但是插入操作是o(n),所以总体可能是o(n^2)

巧妙的解法是利用中卫数的性质， 比它小的或等于它的在数组中的数 和 比它大或等于它的在数组中的数的个数是相等的，或差一在偶数个数的数组中。所以我们可以通过维护两个集合，第一个是所有都比中位数小的，第二个是所有都比中位数大的，如果这两个集合中元素个数不等，则往少的那边转移，因为每次最多进来一个数，所以只用转移一次。

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[] ans = new int[nums.length];
        if (nums.length == 0) {
            return ans;
        }

        PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>();
        PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(new Comparator<Integer>(){
            public int compare(Integer a, Integer b) {
                return b - a;
            }
        });

        int median = nums[0];
        ans[0] = median;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] > median) {
                minHeap.offer(nums[i]);
            } else {
                maxHeap.offer(nums[i]);
            }

            if (maxHeap.size() < minHeap.size() - 1) {
                maxHeap.offer(median);
                median = minHeap.poll();
            } else if (maxHeap.size() > minHeap.size()) {
                minHeap.offer(median);
                median = maxHeap.poll();
            }
            ans[i] = median;
        }
        return ans;

    }
}