[leetcode] 239. Sliding Window Maximum 解题报告

题目链接: https://leetcode.com/problems/sliding-window-maximum/

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?

Hint:

1. How about using a data structure such as deque (double-ended queue)?
2. The queue size need not be the same as the window’s size.
3. Remove redundant elements and the queue should store only elements that need to be considered.

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思路: 利用一个双端队列, 我们可以在O(1)的时间内删除队首和队尾元素, 并且可以在O(1)的时间内随机存取, 也就是说双端队列综合了链表和数组的优点.

在队列中我们存储元素在数组中的位置, 并且维持队列的严格递减, 也就说队首元素最大的,代表在数组中到队首元素那个位置之前最大的数. 

当遍历到一个新元素时, 如果队列里有比当前元素小的, 就将其移除队列, 以保证队列的递减. 

当队列元素位置之差大于k, 就将队首元素移除.

代码如下:

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        if(nums.size() ==0 || k == 0) return {};
        vector<int> result;
        deque<int> que;
        for(int i = 0; i< nums.size(); i++)
        {
            while(!que.empty() && nums[i] > nums[que.back()])
                que.pop_back();
            que.push_back(i);
            if(i >= que.front() + k) que.pop_front();
            if(i >= k-1) result.push_back(nums[que.front()]);
        }
        return result;
    }
};

参考: https://leetcode.com/discuss/74409/a-concise-solution-using-deque