239. Sliding Window Maximum 239.滑动窗口最大值

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.

Follow up:
Could you solve it in linear time?

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

 

思路：起码两个指针的暴力做法可以想一想。

public int[] maxSlidingWindow(int[] nums, int k) {
    if(nums == null || k <= 0) return new int [0];
    int [] arr = new int[nums.length - k + 1];
    for(int i = 0; i < nums.length - k + 1; i++){
        int max = Integer.MIN_VALUE;
        for(int j = i; j < i + k; j++)
           max = Math.max(max, nums[j]);
        arr[i] = max;
    }
    return arr;
}

 coinbase oa

public int minSlidingWindow(List<Integer> nums, int k) {
    if(nums == null || k <= 0) return -1;
    int [] arr = new int[nums.size() - k + 1];
    for(int i = 0; i < nums.size() - k + 1; i++){
        int min = Integer.MAX_VALUE;
        for(int j = i; j < i + k; j++)
           min = Math.min(min, nums.get(j));
        arr[i] = min;
    }

    int max = Integer.MIN_VALUE;
    for (int k = 0; k < arr.length; k++) {
        max = Math.max(max, arr[k]);
    }
    return max;
}

 能全过的代码

    public static int segment(int w, List<Integer> in) {
    // Write your code here
    final int[] min_left = new int[in.size()];
final int[] min_right = new int[in.size()];

    min_left[0] = in.get(0);
    min_right[in.size() - 1] = in.get(in.size() - 1);

    for (int i = 1; i < in.size(); i++) {
        min_left[i] = (i % w == 0) ? in.get(i) : Math.min(min_left[i - 1], in.get(i));

        final int j = in.size() - i - 1;
        min_right[j] = (j % w == 0) ? in.get(j) : Math.min(min_right[j + 1], in.get(j));
    }

    final int[] sliding_min = new int[in.size() - w + 1];
    for (int i = 0, j = 0; i + w <= in.size(); i++) {
        sliding_min[j++] = Math.min(min_right[i], min_left[i + w - 1]);
    }

    int max = Integer.MIN_VALUE;
    for (int idx = 0; idx < sliding_min.length; idx++) {
        max = Math.max(max, sliding_min[idx]);
    }
    return max;
    }

}