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:
- 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.
参考fentoyal的单调队列做法，我加上一些解释。
双端队列mq存放的是二元组，
第一元素表示当前值，
第二元素表示在队列中末尾值之后，在当前值之前并且小于当前值的个数。
mq的首个二元组的第一元素表示窗口范围内的最大值。
max函数：
返回首个二元组的第一元素
push函数：
为了保证max函数的作用，需要从末尾开始，把小于当前值的连续二元组出队列，但是需要累加上小于当前值的数目。
pop函数：
如果窗口在首个二元组之前还有元素，那么对队列影响就是首个二元组的计数减一。
否则代表首个二元组就是待出队的值，出队。
class MonoQue
{
public:
deque<pair<int, int> > q;
int maxV()
{
return q.front().first;
}
void push(int n)
{
int count = ;
while(!q.empty() && q.back().first < n)
{
count += (q.back().second + );
q.pop_back();
}
q.push_back(make_pair(n, count));
}
void pop()
{
if(q.front().second > )
q.front().second --;
else
q.pop_front();
}
}; class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
vector<int> ret;
if(k == )
return ret;
MonoQue mq;
for(int i = ; i < k; i ++)
mq.push(nums[i]);
for(int i = k; i < nums.size(); i ++)
{
ret.push_back(mq.maxV());
mq.pop();
mq.push(nums[i]);
}
ret.push_back(mq.maxV());
return ret;
}
};