leetcode No239. Sliding Window Maximum
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Question:
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.
Algorithm:
用双向队列来解决问题
Ex:1 3 -1 -3 5 3 6 7 size=3
1、因为此时队列为空,在队尾插入1
back 1(下标0) front
2、因为3比1大,所以1不可能是最大的元素,所以把1删除,pop_back(),并在队尾插入1
back 3(下标1) front
3、-1比3小,如果3删除后,-1可能是最大的元素,所以在队尾插入-1
back -1(下标2) 3(下标1) front
4、-3比-1小,所以在队尾插入-3
back -3(下标3) -1(下标2) 3(下标1) front
5、5比-3、-1、3都大,所以-3、-1、3删除,并在队尾插入5
back 5(下标4) front
6、3比5小,在队尾插入3
back 3(下标5) 5(下标4) front
7、6比3、5大,删除3、5,在队尾插入6
back 6(下标6) front
8、7比6大,删除6、在队尾插入7
back 7 front
注意:如果此时的滑动窗口不包括队头元素了,要把队头元素删除
判断滑动窗口是否包含一个数字,应该在队列里存入数字在数组里的下标,而不是数值。但一个数字的下标与当前处理的数字的下标之差大于等于size时,这个数字可以从队列里删除
Accepted Code:
class Solution {public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { vector<int> res; if(nums.empty() || k==0 || k>nums.size()) return res; deque<int> d; int i=0; for(;i<k;i++){ while(!d.empty() && nums[i]>=nums[d.back()]) d.pop_back(); d.push_back(i); } for(;i<nums.size();i++){ res.push_back(nums[d.front()]); while(!d.empty() && nums[i]>=nums[d.back()]) d.pop_back(); if(!d.empty() && k<=(i-d.front())) d.pop_front(); d.push_back(i); } res.push_back(nums[d.front()]); return res; }};
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