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

Solution, using deque

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

Solution, using heap

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> res;
        if ( n == 0 ) return res;
        priority_queue<pair<int, int>> heap;
        for ( int i = 0; i < k; i++ ) heap.push(pair<int, int>(nums[i], i));
        res.push_back(heap.top().first);
        for ( int i = k; i < n; i++ ) {
            heap.push(pair<int, int>(nums[i], i));
            while ( ! heap.empty() and heap.top().second <= i-k ) heap.pop();
            res.push_back(heap.top().first);
        }
        return res;
    }
};