239. 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. 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?

---

解法一：

求最大值，可以考虑用heap。可是当sliding window移动时，heap不能找到需要删除的element(only top element is visiable in heap)，to solve this issue, we can use a hash table to store the elements of the sliding window: when calculating the max value, check whether the heap's top element is in the hash map or not, if in, it is the max value; otherwise, pop the top and continue until find the max value.

时间复杂度：this is not a linear time algorithm. 

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        priority_queue<int> pq;     //easily to find the max value
        unordered_map<int,int> map; //store nums in the sliding window
        vector<int> result;
        if(nums.size()==0){
            return result;
        }

        for(int i=0;i<k;i++){
            pq.push(nums[i]);
            map[nums[i]]++;
        }
        result.push_back(pq.top());
        for(int i=k;i<nums.size();i++){
            pq.push(nums[i]);
            map[nums[i]]++;
            map[nums[i-k]]--;
            if(map[nums[i-k]]==0){
                map.erase(nums[i-k]);
            }
            
            //priority queue cannot remove a value easily, we can check whether its top value is in the map
            while(map.find(pq.top())==map.end()){ 
                pq.pop();
            }
            result.push_back(pq.top());
        }
        return result;
    }
};

 

解法二：linear algorithm

使用双向队列。队列存储数组下标，里面的元素是最大值的下标

 

参考答案：

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> result;
        deque<int> q; //sliding window's possible max values
        for (int i = 0; i < nums.size(); ++i) {
            if (!q.empty() && q.front() == i - k) q.pop_front();
            
            //if current element gets in, elements in dequeue that is smaller need to be removed  
            while (!q.empty() && nums[q.back()] < nums[i]) q.pop_back(); 
            q.push_back(i);
            if (i >= k - 1) result.push_back(nums[q.front()]);
        }
        return result;
    }
};