leetcode 239: Sliding Window Maximum

最新推荐文章于 2020-01-03 03:40:56 发布

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本文详细介绍了如何使用双端队列（deque）来解决滑动窗口内的最大值问题，通过移除旧元素并添加新元素保持队列的单调递减特性，确保每次窗口移动后都能快速获取当前窗口的最大值。

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Use the a deque to save numbers like a waitlist. If the window size equals to k, start the pop operation of the deque and push the front number into the res vector. If nums[i-k] equals to the front number of deque, the front should be popped because the window moves. Then a new number nums[i] should be pushed from back of the deque. While pushing, all numbers at the back of the deque which are smaller than nums[i] should be removed because they came earlier than nums[i] and if they are moved out of the window later, all numbers that are bigger than them are still in the deque, so the maximum number won't be affected.

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