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239.Sliding-Window-Maximum

239. Sliding Window Maximum

题目地址

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:
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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;
}
}