239. Sliding Window Maximum

用Dequeue

 

 

 1 class Solution {
 2     public int[] maxSlidingWindow(int[] nums, int k) {
 3         if(nums.length == 0 || nums.length < k) return new int[0];
 4         int[] res = new int[nums.length - k + 1];
 5         Deque<Integer> deque = new LinkedList<>();
 6         for(int i = 0; i < nums.length; i++){
 7             if(!deque.isEmpty() && deque.peek() < i - k + 1){
 8                 deque.poll();
 9             }
10             while(!deque.isEmpty() && nums[deque.peekLast()] < nums[i]){
11                 deque.pollLast();
12             }
13             deque.offer(i);
14             if(i >= k-1){
15                 res[i - k + 1] = nums[deque.peek()];
16             }
17         }
18         return res;
19         
20     }
21 }
22 
23 
24 
25 
26 //O(n^2)
27 class Solution {
28     public int[] maxSlidingWindow(int[] nums, int k) {
29         int[] res = new int[nums.length-k+1];
30         if(nums.length == 0 || nums.length < k) return new int[]{};
31         int begin = 0, end = k-1;
32         int max = Integer.MIN_VALUE;
33         for(int i = begin; i <= end; i++){
34              max = Math.max(nums[i], max);   
35         }
36         res[begin] = max;
37         begin++;
38         end++;
39         while(end < nums.length){
40             res[begin] = Integer.MIN_VALUE;
41             for(int i = begin; i <= end; i++){
42                 res[begin] = Math.max(res[begin], nums[i]);
43             }
44             begin++;
45             end++;
46         }
47         return res;
48         
49     }
50 }

 

posted @ 2018-10-22 06:00  jasoncool1  阅读(110)  评论(0)  编辑  收藏  举报