239. Sliding Window Maximum
2021-05-17 23:31
标签：xpl int min input out div array new integer 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: Note: Follow up: 239. Sliding Window Maximum 标签：xpl int min input out div array new integer 原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11757862.htmlInput: 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
You may assume k is always valid, 1 ≤ k ≤ input array‘s size for non-empty array.
Could you solve it in linear time?class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if(nums.length == 0) return new int[]{};
int[] res = new int[nums.length - k + 1];
for(int i = 0; i ){
int tmp = Integer.MIN_VALUE;
for(int j = i; j ){
tmp = Math.max(tmp, nums[j]);
}
res[i] = tmp;
}
return res;
}
}