LeetCode OJ 之 Sliding Window Maximum(滑動視窗的最大值)
阿新 • • 發佈:2019-02-15
題目:
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?
思路:
這裡可以使用雙端佇列,使得佇列裡的數呈遞減趨勢,即始終保持隊頭最大。
如果對頭在當前視窗內,則它就是當前視窗的最大值。
如果隊頭不在當前視窗範圍,則刪除它。這裡佇列裡儲存的不是值,而是值的位置,因為如果儲存值的話,無法判斷值是否在當前視窗中。
程式碼:
class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { deque<int> dq; vector<int> result; int len = nums.size(); for (int i = 0 ; i < len ; i++) { //當前視窗是[i-k+1, i-k+2, ... i-1, i ](i>=k-1時),如果隊頭不在這個範圍內,則丟棄 if (!dq.empty() && dq.front() < i-k+1) dq.pop_front(); //把佇列中小的刪掉,保持佇列遞減 while (!dq.empty() && nums[dq.back()] < nums[i]) dq.pop_back(); dq.push_back(i); //從k-1開始,每次的隊頭就是最大值 if (i >= k-1) result.push_back(nums[dq.front()]); } return result; } };
上面deque的操作list也都有,可以替換成list。