Sliding Window Maximum
问题描述
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.
For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
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
Therefore, return the max sliding window as [3,3,5,5,6,7].
解决思路
1. 最直观的方法就是逐个比较，时间复杂度为O(kn);
2. 另一种更加巧妙的方法是借助一个双向队列，滑窗的同时记录下当前窗口的最大值。
具体做法为
1. 输入元素的个数不足k时，进队列；
2. 否则，比较当前元素和队列尾部的元素，如果队列尾部的元素小于当前元素则不断地将队尾元素出队；
3. 每次记录下队列的头元素为滑动窗口中的最大元素，并且需要判断该最大元素是否为窗口的首元素，如果是则需要移除。
程序
public class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || nums.length == 0 || nums.length < k) {
return new int[0];
}
LinkedList<Integer> doublyQueue = new LinkedList<Integer>();
int[] maxs = new int[nums.length - k + 1];
for (int i = 0; i < nums.length; i++) {
while (!doublyQueue.isEmpty() && doublyQueue.getLast() < nums[i]) {
doublyQueue.removeLast();
}
doublyQueue.add(nums[i]);
if (i < k - 1) {
continue;
}
maxs[i - k + 1] = doublyQueue.getFirst();
if (doublyQueue.getFirst() == nums[i - k + 1]) {
doublyQueue.removeFirst();
}
}
return maxs;
}
}