Link: https://leetcode.cn/problems/sliding-window-maximum/
Question
difficulty: high
adj diff: 4
You are given an array of integers 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 1:
Input: nums = [1,3,-1,-3,5,3,6,7], 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
Example 2:
Input: nums = [1], k = 1
Output: [1]
Constraints:
1 <= nums.length <= 105
-104 <= nums[i] <= 104
1 <= k <= nums.length
这道题,只要好好写,注意边界就可以了。
单调递减队列问题。
注意的就是 = 的情况。相等的元素,保留在 list 中,一个一个地 remove,没问题。
Code
public int[] maxSlidingWindow(int[] nums, int k) {
int len = nums.length;
int[] ans = new int[len + 1 - k];
List<Integer> list = new LinkedList<Integer>();
// init the windows of size k elements
for (int i = 0; i < k; i++) {
while (list.size() > 0 && list.get(list.size() - 1) < nums[i]) {
list.remove(list.size() - 1);
}
list.add(nums[i]);
}
for (int i = k; i < len; i++) {
ans[i - k] = list.get(0);
// remove nums[i-k+1], add nums[i]
if (list.get(0) == nums[i - k]) {
list.remove(0);
}
while (list.size() > 0 && list.get(list.size() - 1) < nums[i]) {
list.remove(list.size() - 1);
}
list.add(nums[i]);
}
ans[len - k] = list.get(0);
return ans;
}