leetcode——Sliding Window Maximum

题目

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].

Note:

You may assume k is always valid, ie: 1 ≤ k ≤ input array’s size for non-empty array.

Follow up:

Could you solve it in linear time?

思路

其实这是一道数据结构的题，需要用到双端队列。其实队列的长度上限只需要设置为窗口大小即可。这个双端队列中只需要存，从开始数字向右递增的数字：比如，第一个窗口【1，3，-1】，队列中只需要记录1和3，当窗口右移一个的时候，队列出队剩下3，此时窗口【3，-1，-3】，同样-3不入队。再当窗口右移一个时，队列为空，此时窗口【-1，-3，5】，当对列空的时候，需要从窗口左侧开始将递增的数字依次入队形成新的队列，此时队列中为-1，5.再向右移动，队列出队剩下5，窗口为【-3，5，3】，3不需要入队。再右移，队列成空，此时窗口为【5，3，6】，队列重置为5，6；以此类推，因为每当队列变为空的时候，都要从当前窗口重新加载，因此不会忽略掉任何一个最大值。

代码

public int[] maxSlidingWindow(int[] nums, int k) {
        ArrayDeque<Integer> dequeue = new ArrayDeque<Integer>();
        List<Integer> arr = new LinkedList<Integer>();
        int curIdx = 0;

        while(curIdx!=nums.length){
            if(dequeue.isEmpty()){
                for(int i=0;i<k && curIdx<nums.length;i++){
                    if(dequeue.isEmpty()){
                        dequeue.add(nums[curIdx]);
                    } else{
                        if(nums[curIdx]>dequeue.peekLast()){
                            dequeue.add(nums[curIdx]);
                        }
                    }
                    curIdx++;
                }
                arr.add(dequeue.peekLast());
            } else{
                dequeue.pollFirst();
                if(dequeue.isEmpty()){
                    for(int i=curIdx-k+1;i<=curIdx;i++){
                        if(dequeue.isEmpty()){
                            dequeue.add(nums[i]);
                        } else{
                            if(nums[i]>dequeue.peekLast()){
                                dequeue.add(nums[i]);
                            }
                        }
                    }
                } else{
                    if(nums[curIdx]>dequeue.peekLast()){
                        dequeue.add(nums[curIdx]);
                    }
                }

                arr.add(dequeue.peekLast());
                curIdx++;
            }
        }
        int [] rst = new int[arr.size()];
        int i=0;
        for(Integer I:arr){
            rst[i] = I;
            i++;
        }

        return rst;
    }

版权声明：本文为博主原创文章，未经博主允许不得转载。