这个函数的时间复杂度是多少？

Question

这是 Java 中 Sliding Window Maximum 问题的示例解决方案。

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.

我想获取此函数的时间和 space 复杂度。这是我认为的答案：

时间：O((n-k)(k * logk)) == O(nklogk)

Space（辅助）：O(n) 代表 return int[] 和 O(k) 代表 pq。总计 O(n).

这是正确的吗？

private static int[] maxSlidingWindow(int[] a, int k) {
    if(a == null || a.length == 0) return new int[] {};
    PriorityQueue<Integer> pq = new PriorityQueue<Integer>(k, new Comparator<Integer>() {
        // max heap
        public int compare(Integer o1, Integer o2) {
            return o2 - o1;
        }
    });
    int[] result = new int[a.length - k + 1];
    int count = 0;
    // time: n - k times
    for (int i = 0; i < a.length - k + 1; i++) {
        for (int j = i; j < i + k; j++) {
            // time k*logk (the part I'm not sure about)
            pq.offer(a[j]);
        }

        // logk
        result[count] = pq.poll();
        count = count + 1;
        pq.clear();
    }
    return result;
}

Answer 1

除了-

，你大部分都是对的

for (int j = i; j < i + k; j++) {
     // time k*logk (the part I'm not sure about)
     pq.offer(a[j]);
}

此处总执行次数为log1 + log2 + log3 + log4 + ... + logk。这个系列的总和-

log1 + log2 + log3 + log4 + ... + logk = log(k!)

第二个想法是，你可以使用双端队列属性比你的线性时间解决方案做得更好，这将是 O(n)。这是我的解决方案 -

public int[] maxSlidingWindow(int[] nums, int k) {      
    if (nums == null || k <= 0) {
        return new int[0];
    }
    int n = nums.length;
    int[] result = new int[n - k + 1];
    int indx = 0;

    Deque<Integer> q = new ArrayDeque<>();

    for (int i = 0; i < n; i++) {

        // remove numbers out of range k
        while (!q.isEmpty() && q.peek() < i - k + 1) {
            q.poll();
        }

        // remove smaller numbers in k range as they are useless
        while (!q.isEmpty() && nums[q.peekLast()] < nums[i]) {
            q.pollLast();
        }

        q.offer(i);
        if (i >= k - 1) {
            result[indx++] = nums[q.peek()];
        }
    }

    return result;
}

HTH.

这个函数的时间复杂度是多少？

What is the time complexity of this function?

big-o

time-complexity