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 is

A[(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;
}
}