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- 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?
复制代码
- class Solution {
- public int[] maxSlidingWindow(int[] nums, int k) {
- int i = 0, j;
- int n = nums.length;
- List <Integer> list = new ArrayList<>();
- List <Integer> max = new ArrayList<>();
- for(j = 0; j < n ; j++){
- max.add(nums[j]);
-
- if(j - i + 1>= k){
- list.add(Collections.max(max));
- max.remove(i);
- }
- }
-
- int[] ret = list.stream().mapToInt(x ->x).toArray();
- return ret;
- }
- }
复制代码
- class Solution {
- public int[] maxSlidingWindow(int[] nums, int k) {
- if(nums.length == 0 || k == 0) return nums;
- int j = 0, n = nums.length;
- int[] arr = new int[n-k+1];
- for(int i = 0; i< n ; i++){
- int max = Integer.MIN_VALUE;
- if(i - j +1 >= k){
- for(int m = j; m <= i; m++){
- max = Math.max(max, nums[m]);
- }
- arr[i-k+1] = max;
- j++;
-
- }
- }
-
- return arr;
- }
- }
复制代码
monotonic queue!
- class Solution {
- public int[] maxSlidingWindow(int[] nums, int k) {
- if(nums.length == 0) return nums;
- int n = nums.length;
- int[] res = new int[n - k +1];
- ArrayDeque<Integer> index = new ArrayDeque<>();
- for(int i =0; i< n; i++){
- while(!index.isEmpty() && nums[i] > nums[index.peekLast()]){
- index.pollLast();
- }
- index.offerLast(i);
- if(i - k + 1 >= 0){
- res[i - k+1] = nums[index.peekFirst()];
- }
- if(i - k + 1>= index.peekFirst()) index.pop();
- }
-
- return res;
- }
- }
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