leetcode 239. Sliding Window Maximum

最新推荐文章于 2022-09-13 10:47:21 发布

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最新推荐文章于 2022-09-13 10:47:21 发布

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本文介绍了一种解决滑动窗口最大值问题的有效算法。通过使用双端队列存储递减序列的方式，确保每次滑动窗口移动时都能快速找到当前窗口内的最大值。此方法能在O(n)的时间复杂度内解决问题。

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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?

好像剑指offer有这个题。

运用一个队列，里面只存递减的数。

从尾部进来一个数，把前面比它小的先弹出去，保证递减性。这样最大值一直在队列头部。

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