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[LeetCode] 239. Sliding Window Maximum #239

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grandyang opened this issue May 30, 2019 · 0 comments
Open

[LeetCode] 239. Sliding Window Maximum #239

grandyang opened this issue May 30, 2019 · 0 comments

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@grandyang
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grandyang commented May 30, 2019

 

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?

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.

 

这道题给定了一个数组,还给了一个窗口大小k,让我们每次向右滑动一个数字,每次返回窗口内的数字的最大值。难点就在于如何找出滑动窗口内的最大值(这不废话么,求得不就是这个),那么最狂野粗暴的方法就是每次遍历窗口,找最大值呗,OJ 说呵呵哒,no way!我们希望窗口内的数字是有序的,但是每次给新窗口排序又太费时了,所以最好能有一种类似二叉搜索树的结构,可以在 lgn 的时间复杂度内完成插入和删除操作,那么使用 STL 自带的 multiset 就能满足我们的需求,这是一种基于红黑树的数据结构,可以自动对元素进行排序,又允许有重复值,完美契合。所以我们的思路就是,遍历每个数字,即窗口右移,若超过了k,则需要把左边界值删除,这里不能直接删除 nums[i-k],因为集合中可能有重复数字,我们只想删除一个,而 erase 默认是将所有和目标值相同的元素都删掉,所以我们只能提供一个 iterator,代表一个确定的删除位置,先通过 find() 函数找到左边界 nums[i-k] 在集合中的位置,再删除即可。然后将当前数字插入到集合中,此时看若 i >= k-1,说明窗口大小正好是k,就需要将最大值加入结果 res 中,而由于 multiset 是按升序排列的,最大值在最后一个元素,我们可以通过 rbeng() 来取出,参见代码如下:

 

解法一:

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> res;
        multiset<int> st;
        for (int i = 0; i < nums.size(); ++i) {
            if (i >= k) st.erase(st.find(nums[i - k]));
            st.insert(nums[i]);
            if (i >= k - 1) res.push_back(*st.rbegin());
        }
        return res;
    }
};

 

我们也可以使用优先队列来做,即最大堆,不过此时我们里面放一个 pair 对儿,由数字和其所在位置组成的,这样我们就可以知道每个数字的位置了,而不用再进行搜索了。在遍历每个数字时,进行 while 循环,假如优先队列中最大的数字此时不在窗口中了,就要移除,判断方法就是将队首元素的 pair 对儿中的 second(位置坐标)跟 i-k 对比,小于等于就移除。然后将当前数字和其位置组成 pair 对儿加入优先队列中。此时看若 i >= k-1,说明窗口大小正好是k,就将最大值加入结果 res 中即可,参见代码如下:

 

解法二:

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> res;
        priority_queue<pair<int, int>> q;
        for (int i = 0; i < nums.size(); ++i) {
            while (!q.empty() && q.top().second <= i - k) q.pop();
            q.push({nums[i], i});
            if (i >= k - 1) res.push_back(q.top().first);
        }
        return res;
    }
};

 

题目中的 Follow up 要求我们代码的时间复杂度为 O(n)。提示我们要用双向队列 deque 来解题,并提示我们窗口中只留下有用的值,没用的全移除掉。果然 Hard 的题目我就是不会做,网上看到了别人的解法才明白,解法又巧妙有简洁,膜拜啊。大概思路是用双向队列保存数字的下标,遍历整个数组,如果此时队列的首元素是 i-k 的话,表示此时窗口向右移了一步,则移除队首元素。然后比较队尾元素和将要进来的值,如果小的话就都移除,然后此时我们把队首元素加入结果中即可,参见代码如下:

 

解法三:

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

 

类似题目:

Minimum Window Subsequence

Min Stack

Longest Substring with At Most Two Distinct Characters

Paint House II

 

参考资料:

https://leetcode.com/problems/sliding-window-maximum/

https://leetcode.com/problems/sliding-window-maximum/discuss/65936/My-Java-Solution-Using-PriorityQueue

https://leetcode.com/problems/sliding-window-maximum/discuss/65884/Java-O(n)-solution-using-deque-with-explanation

 

LeetCode All in One 题目讲解汇总(持续更新中...)

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