Description
Numbers keep coming, return the median of numbers at every time a new number added.
Clarification
What's the definition of Median?
- The
median
is not equal tomedian
in math. - Median is the number that in the middle of a sorted array. If there are n numbers in a sorted array A, the median is A[(n - 1) / 2]A[(n−1)/2].
- For example, if
A=[1,2,3]
, median is2
. IfA=[1,19]
, median is1
.
Example
Example 1
Input: [1,2,3,4,5]
Output: [1,1,2,2,3]
Explanation:
The medium of [1] and [1,2] is 1.
The medium of [1,2,3] and [1,2,3,4] is 2.
The medium of [1,2,3,4,5] is 3.
Example 2
Input: [4,5,1,3,2,6,0]
Output: [4,4,4,3,3,3,3]
Explanation:
The medium of [4], [4,5], [4,5,1] is 4.
The medium of [4,5,1,3], [4,5,1,3,2], [4,5,1,3,2,6] and [4,5,1,3,2,6,0] is 3.
Challenge
Total run time in O(nlogn).
思路:用 maxheap 保存左半部分的数,用 minheap 保存右半部分的数。
把所有的数一左一右的加入到每个部分。左边部分最大的数就一直都是 median。
这个过程中,可能会出现左边部分并不完全都 <= 右边部分的情况。这种情况发生的时候,交换左边最大和右边最小的数即可。
public class Solution { /** * @param nums: A list of integers. * @return: the median of numbers */ private PriorityQueue<Integer> maxHeap, minHeap; private int numOfElements = 0; public int[] medianII(int[] nums) { // write your code here Comparator<Integer> revCmp = new Comparator<Integer>() { @Override public int compare(Integer left, Integer right) { return right.compareTo(left); } }; int cnt = nums.length; maxHeap = new PriorityQueue<Integer>(cnt, revCmp); minHeap = new PriorityQueue<Integer>(cnt); int[] ans = new int[cnt]; for (int i = 0; i < cnt; ++i) { addNumber(nums[i]); ans[i] = getMedian(); } return ans; } void addNumber(int value) { maxHeap.add(value); if (numOfElements%2 == 0) { if (minHeap.isEmpty()) { numOfElements++; return; } else if (maxHeap.peek() > minHeap.peek()) { Integer maxHeapRoot = maxHeap.poll(); Integer minHeapRoot = minHeap.poll(); maxHeap.add(minHeapRoot); minHeap.add(maxHeapRoot); } } else { minHeap.add(maxHeap.poll()); } numOfElements++; } int getMedian() { return maxHeap.peek(); } }