Data Stream Median
Data Stream Median (leetcodelintcode)
Description
Numbers keep coming, return the median of numbers at every time a new number added.
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.
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).
解题思路
一、优先队列/堆
- 使用两个堆
Heap和一个变量median，median存储当前的中位数，最大堆maxHeap存储小于median的数，最小堆minHeap存储大于median的数。 - 新插入数值比
median小，放入maxHeap；比median大，放入minHeap。 - 比较
maxHeap和minHeap的大小，本解法中其实是比较maxHeap.size() + 1和minHeap的大小。- 如果
maxHeap.size()>minHeap.size()，那么将median放入minHeap，取出maxHeap最大值作为当前median。 - 如果
maxHeap.size() + 1<minHeap.size()，那么将median放入maxHeap，取出minHeap最小值作为当前median。
- 如果
Java 中的PriorityQueue默认是最小堆，堆顶元素是最小值，所以处理最大堆时取了负数。
Java 实现
public
class
Solution
{
/**
*
@param
nums: A list of integers.
*
@return
: the median of numbers
*/
public
int
[] medianII(
int
[] nums) {
int
[] results =
new
int
[nums.length];
if
(nums ==
null
|| nums.length ==
0
) {
return
results;
}
int
median = nums[
0
];
results[
0
] = median;
PriorityQueue
<
Integer
>
maxHeap =
new
PriorityQueue
<
Integer
>
();
PriorityQueue
<
Integer
>
minHeap =
new
PriorityQueue
<
Integer
>
();
for
(
int
i =
1
; i
<
nums.length; i++) {
if
(nums[i]
<
median) {
maxHeap.add(-nums[i]);
}
else
{
minHeap.add(nums[i]);
}
if
(maxHeap.size()
>
minHeap.size()) {
minHeap.add(median);
median = -maxHeap.peek();
maxHeap.remove();
}
else
if
(maxHeap.size() +
1
<
minHeap.size()) {
maxHeap.add(-median);
median = minHeap.peek();
minHeap.remove();
}
results[i] = median;
}
return
results;
}
}
二、优先队列/堆 II
和方法一的实现思路类似，具体实现细节有所不同。不再单独使用变量存储 median ，维持 maxHeap 和 minHeap 的大小，maxHeap 可以比 minHeap 多一个元素， maxHeap 堆顶即为中位数 median 。
本方法的实现更加直观。
Java 实现
public
class
Solution
{
/**
*
@param
nums: A list of integers.
*
@return
: the median of numbers
*/
public
int
[] medianII(
int
[] nums) {
int
[] results =
new
int
[nums.length];
if
(nums ==
null
|| nums.length ==
0
) {
return
results;
}
Comparator
<
Integer
>
revComp =
new
Comparator
<
Integer
>
() {
public
int
compare
(Integer left, Integer right)
{
return
right - left;
}
};
PriorityQueue
<
Integer
>
minHeap =
new
PriorityQueue
<
Integer
>
(nums.length);
PriorityQueue
<
Integer
>
maxHeap =
new
PriorityQueue
<
Integer
>
(nums.length, revComp);
results[
0
] = nums[
0
];
maxHeap.add(nums[
0
]);
for
(
int
i =
1
; i
<
nums.length; i++) {
maxHeap.add(nums[i]);
minHeap.add(maxHeap.poll());
if
(minHeap.size()
>
maxHeap.size()) {
maxHeap.add(minHeap.poll());
}
results[i] = maxHeap.peek();
}
return
results;
}
}