Sliding Window Median
Given an array of n integer, and a moving window(size k), move the window at each iteration from the start of the array, find the median of the element inside the window at each moving. (If there are even numbers in the array, return the N/2-th number after sorting the element in the window. )
For array [1,2,7,8,5], moving window size k = 3. return [2,7,7]
At first the window is at the start of the array like this
[ | 1,2,7 | ,8,5] , return the median 2;
then the window move one step forward.
[1, | 2,7,8 | ,5], return the median 7;
then the window move one step forward again.
[1,2, | 7,8,5 | ], return the median 7;
O(nlog(n)) time
Solution: 用分别用两个heap来维护这个median结构，左边一个maxHeap，右边一个minHeap，在加上当中一个median数值，两个Heap的size差最多为1，即leftSize==rightSize||leftSize+1==rightSize.
首先是要先初始化这个结构，把前k个数放进去，接下里window每向右移一格，等于是加上第nums[i+k-1] 然后 remove nums[i-1]。
注意的是：java里有heap的实现，可用PriorityQueue然后可以定义Comparator, 默认是minHeap, 如果要用maxHeap要自己写一个comparator传入constructor. 但是PriorityQueue的remove方法是O(n)的 即每次remove需要遍历一遍这个Heap找到那个数再Remove，面试时可以和面试官讨论，这一步可以用HashHeap来做，这样remove变成了O(logn), (找到这个数是O(1), remove相当于和最后一个数swap 然后siftdown操作所以是O(logn)).
如果用HashHeap的话，整个Time Complexity就是O(nlogn).
public ArrayList<Integer> medianSlidingWindow(int[] nums, int k) {
ArrayList<Integer> result = new ArrayList<>();
if (nums == null || nums.length == 0 || nums.length < k) {
return result;
}
PriorityQueue<Integer> leftMaxHeap = new PriorityQueue<>(1, new Comparator<Integer>() {
@Override
public int compare(Integer a, Integer b) {
if (a < b) {
return 1;
} else if (a == b) {
return 0;
} else {
return -1;
}
}
});
PriorityQueue<Integer> rightMinHeap = new PriorityQueue<>();
int median = nums[0];
//init the two heaps for the first k elements
for (int i = 1; i < k; i++) {
median = addElement(nums, i, median, leftMaxHeap, rightMinHeap);
}
result.add(median);
//start moving the window forward
for (int i = 1; i <= nums.length - k; i++) {
//add nums[i+k-1]
median = addElement(nums, i + k - 1, median, leftMaxHeap, rightMinHeap);
//remove nums[i-1]
median = removeElement(nums, i - 1, median, leftMaxHeap, rightMinHeap);
result.add(median);
}
return result;
}
public int addElement(int[] nums, int index, int median,
PriorityQueue<Integer> leftMaxHeap, PriorityQueue<Integer> rightMinHeap) {
//the left heap can only be the same size as the right heap
//or left size + 1 = right size
if (leftMaxHeap.size() == rightMinHeap.size()) {
if (nums[index] < median) {
leftMaxHeap.offer(nums[index]);
rightMinHeap.offer(median);
median = leftMaxHeap.poll();
} else {
rightMinHeap.offer(nums[index]);
}
} else {
if (nums[index] <= median) {
leftMaxHeap.offer(nums[index]);
} else {
leftMaxHeap.offer(median);
rightMinHeap.offer(nums[index]);
median = rightMinHeap.poll();
}
}
return median;
}
public int removeElement(int[] nums, int index, int median,
PriorityQueue<Integer> leftMaxHeap, PriorityQueue<Integer> rightMinHeap) {
//the left heap can only be the same size as the right heap
//or left size + 1 = right size
if (leftMaxHeap.size() == rightMinHeap.size()) {
if (nums[index] == median) {
median = leftMaxHeap.poll();
} else if (nums[index] < median) {
leftMaxHeap.remove(nums[index]);
} else {
rightMinHeap.remove(nums[index]);
if (!leftMaxHeap.isEmpty()) {
rightMinHeap.offer(median);
median = leftMaxHeap.poll();
}
}
} else {
if (nums[index] == median) {
median = rightMinHeap.poll();
} else if (nums[index] <= median) {
leftMaxHeap.remove(nums[index]);
if (!rightMinHeap.isEmpty()) {
leftMaxHeap.offer(median);
median = rightMinHeap.poll();
}
} else {
rightMinHeap.remove(nums[index]);
}
}
return median;
}
