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480. Sliding Window Median
Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value. Examples:
[2,3,4] , the median is 3
[2,3], the median is (2 + 3) / 2 = 2.5
Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Your job is to output the median array for each window in the original array.
For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
Window position Median
--------------- -----
[1 3 -1] -3 5 3 6 7 1
1 [3 -1 -3] 5 3 6 7 -1
1 3 [-1 -3 5] 3 6 7 -1
1 3 -1 [-3 5 3] 6 7 3
1 3 -1 -3 [5 3 6] 7 5
1 3 -1 -3 5 [3 6 7] 6Therefore, return the median sliding window as [1,-1,-1,3,5,6].
*Note: *
You may assume k is always valid, ie: k is always smaller than input array’s size for non-empty array.
Answers within 10^-5 of the actual value will be accepted as correct.
思路分析
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/**
* 答案是正确的，超大数组测试时，超时了。
*
* @author D瓜哥 · https://www.diguage.com
* @since 2024-08-30 11:45:50
*/
public double[] medianSlidingWindow(int[] nums, int k) {
// 如果窗口是奇数，则 topSmall 多一个
Queue<Integer> topSmall = new PriorityQueue<>((a, b) -> Integer.compare(a, b));
// 使用 Integer.compare，防止两个最小负数相减时溢出
Queue<Integer> topLarge = new PriorityQueue<>((a, b) -> Integer.compare(b, a));
double[] result = new double[nums.length - k + 1];
for (int i = 0; i < k; i++) {
topSmall.add(nums[i]);
}
for (int i = 0; i < k / 2; i++) {
topLarge.add(topSmall.poll());
}
// 先相处再相加，防止溢出
result[0] = k % 2 == 1 ? 1.0 * topSmall.peek()
: topSmall.peek() / 2.0 + topLarge.peek() / 2.0;
for (int i = k; i < nums.length; i++) {
int num = nums[i];
if (topSmall.peek() <= num) {
topSmall.add(num);
} else {
topLarge.add(num);
}
int delNum = nums[i - k];
if (topSmall.peek() <= delNum) {
topSmall.remove(delNum);
} else {
topLarge.remove(delNum);
}
// 添加，删除，再平衡双堆，这样才能保证堆的平衡性
// topSmall 应该始终大于等于 topLarge
while (topLarge.size() > topSmall.size()) {
topSmall.add(topLarge.poll());
}
// topSmall 与 topLarge 的差值不能大于 1
while (topSmall.size() - topLarge.size() > 1) {
topLarge.add(topSmall.poll());
}
result[i - k + 1] = k % 2 == 1 ? 1.0 * topSmall.peek()
: topSmall.peek() / 2.0 + topLarge.peek() / 2.0;
}
return result;
}