239.Sliding-Window-Maximum

239. Sliding Window Maximum

题目地址

https://leetcode.com/problems/sliding-window-maximum/

题目描述

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?

代码

Approach 1: Use a hammer (TLE)

Complexity Analysis

• Time complexity : O(Nk), where N is number of elements in the array.

• Space complexity : O(N-k+1) for an output array.

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
      int n = nums.length;
      if (n * k == 0) return new int[0];

      int[] output = new int[n - k + 1];
      for (int i = 0; i < n - k + 1； i++) {
        int max = Integer.MIN_VALUE;
        for (int j = i; j < i + k; j++) {
          max = Math.max(max, nums[j]);
        }
        output[i] = max;
      }

      return output;
  }
}

Approach 2: Deque

Deque 每次将最大的元素的index放在首位

class Solution {
    ArrayDeque<Integer> deq = new ArrayDeque<Integer>();
    int[] nums;
    public int[] maxSlidingWindow(int[] nums, int k) {
      int n = nums.length;
      if (n * k == 0) return new int[0];
      if (k == 1) return nums;

      this.nums = nums;
      int max_idx = 0;
      for (int i = 0; i < k; i++) {
          cleanDeque(i, k);
          deq.addLast(i);
          if (nums[i] > nums[max_idx]) {
              max_idx = i;
          }
      }

      int[] output = new int[n - k + 1];
      output[0] = nums[max_idx];

      for (int i = k; i < n; i++) {
          cleanDeque(i, k);
          deq.addLast(i);
          output[i - k + 1] = nums[deq.getFirst()];
      }

      return output;
    }

  public void cleanDeque(int i, int k) {
    // remove indexes of elements not from sliding window
    if (!deq.isEmpty() && deq.getFirst() == i - k) {
        deq.removeFirst();
    }
    // remove from deq indexes of all elements
    // which are smaller than current elemnt num[i]
    while (!deq.isEmpty() && nums[i] > nums[deq.getLast()]) {
        deq.removeLast();
    }
  }
}

Approach #3 Dynamic programming

[i % k == 0, ~ i] 中左边block的最大值赋值给 left[i], 边界右

[i, i % k == 0] 中右边block的最大值赋值给 right[i]，边界左

Complexity Analysis

• Time complexity : O(N), since all we do is 3 passes along the array of length N.

• Space complexity : O(N) to keep left and right arrays of length N, and output array of length N - k + 1.

class Solution {
  public int[] maxSlidingWindow(int[] nums, int k) {
    int n = nums.length;
    if (n * k == 0)    return new int[0];
    if (k == 1) return nums;

    int[] left = new int[n];
    left[0] = nums[0];
    int[] right = new int[n];
    right[n - 1] = nums[n - 1];
    for (int i = 1; i < n; i++) {
      if (i % k == 0) { 
        left[i] = nums[i];
      } else {
        left[i] = Math.max(left[i - 1], nums[i]);
      }

      int j = n - i - 1;
      if ((j + 1) % k == 0) {
        right[j] = nums[j];
      } else {
        right[j] = Math.max(right[j + 1], nums[j]);
      }

      int[] output = new int[n - k + 1];
      for (int i = 0; i < n - k + 1; i++) {
        output[i] = Math.max(left[i + k - 1], right[i]);
      }

      return output;
  }
}