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. Return the max sliding window.
Example:
Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7]
Explanation:
Window position Max
[1 3 -1] -3 5 3 6 7 3
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 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Note:
You may assume k is always valid, 1 ≤ k ≤ input array's size for non-empty array.
Follow up:
Could you solve it in linear time?
Again 先上暴力解法。雙層嵌套的循環，時間復雜度是O(N*k)=O(N)
class Solution {
/**
* Brute force: every time traverse the whole window to find the maximum. Time complexity: O(N*k)
**/
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0 || nums.length < k) {
return new int[0];
}
int[] maxs = new int[nums.length - k + 1];
for (int i = 0; i < nums.length - k + 1; i++) {
int tempMax = Integer.MIN_VALUE;
for (int j = 0; j < k; j++) {
tempMax = Math.max(tempMax, nums[i + j]);
}
maxs[i] = tempMax;
}
return maxs;
}
}
我一開始沒有想到要使用別的數據結構。但是這道題的tag里還有Heap，我第一時間想到的是PriorityQueue。但發現優先隊列并不好用因為每次對其進行改動之后就要再去查詢一下min & max。看了一下網上比較通用的做法是維護一個單調遞減的隊列，用于存儲備胎可能成為max of the window的數字的index。
從網上找了一個看起來非常簡化的做法（感覺目前自己還沒有辦法獨立寫出這樣的解法）。
class Solution {
/**
* [3,1,-1,-3,5,3,6,7]
* deque={0,1}
* i = 2. deque={0,1,2}, i + 1 == k -> maxs={3}
* i = 3, deque={0,1,2,3}. deque.peek() = 0 < (i - k + 1). deque.pollFirst(). i + 1 > k => maxs={3, nums[1] = 1}
* i = 4, !deque.isEmpty() == false. deque.addLast(i) => deque={4}. maxs={3,1,5}
* i = 5, deque={4}, i - 4 + 1 < k. deque.addLast(i) => deque{4,5} => maxs={3,1,5,5}
* i = 6, deque={4,5} => remove 4 and 5 => deque={6} => maxs={3,1,5,5,6}
* i = 7, deque={6} => remove 6 => deque{7} => maxs={3,1,5,5,6,7}
* Runtime: 11 ms, faster than 59.93% of Java online submissions for Sliding Window Maximum.
* Memory Usage: 41.3 MB, less than 81.25% of Java online submissions for Sliding Window Maximum.
**/
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0 || nums.length < k) {
return new int[0];
}
int[] maxs = new int[nums.length - k + 1];
Deque<Integer> deque = new LinkedList<Integer>();
for (int i = 0; i < nums.length; i++) {
// 如果當前的數字比隊列里之前的數字都要大，就把隊列里這些不可能變成max的數字都移除
while (!deque.isEmpty() && nums[i] > nums[deque.peekLast()]) {
deque.removeLast();
}
while (!deque.isEmpty() && i - deque.peekFirst() + 1 > k) {
deque.pollFirst();
}
deque.addLast(i);
// 由于是從頭開始遍歷，在檢查到第k個元素之前是不會往results里寫入的
if (i + 1 >= k) {
maxs[i + 1 - k] = nums[deque.peekFirst()];
}
}
return maxs;
}
}
More
- 了解Java Deque 的數據結構及其API特性。Java Document: Deque
- 在一篇討論里有人提到了用紅黑樹這樣的BST來存儲備胎。雖然感覺并不必要不過既然提到了，也把紅黑樹這個數據結構放進bucket list吧。Wikipedia: 紅黑樹
