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, 1 ≤ k ≤ input array's size.
Follow up:
Could you solve it in linear time?
Hint:
- How about using a data structure such as deque (double-ended queue)?
- The queue size need not be the same as the window’s size.
- Remove redundant elements and the queue should store only elements that need to be considered.
大意是用雙向隊列可以n的線性時間.網(wǎng)友的解法意思是前端移出了滑動窗口的元素要移除
然后新指向的元素和隊列尾部元素比較，尾部小的元素也要移除.這樣保持隊列總是在滑動窗口里從大到小排好.
個人覺得當k比較大而輸入元素基本隨機時不可能是n復(fù)雜度.而應(yīng)該是(k/2)*n左右
所以我的解法干脆用兩個指針:最大值，第二大值來維護.實際運行還比雙端隊列快一點點.(92ms 擊敗90%)
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
if(nums.size()<2)return nums;
size_t n=nums.size(), maxv=0,secondv=~0;
vector<int> out;
//secondv維持第二大的元素.如果maxv在窗口邊界，secondv就是魔術(shù)~0代表不存在第二小元素.
for(size_t i=1;i<k;++i)
{
if(nums[i]>nums[maxv]){
maxv=i;
secondv=~0;
}else if(secondv==~0||nums[i]>nums[secondv]){
secondv=i;
}
}
out.push_back(nums[maxv]);
for(size_t i=k;i<n;++i)
{
if(maxv<=i-k)
{
if(secondv==~0){
maxv=i;
}else{
maxv=secondv;
secondv=secondv+1;
//maxv移出滑動窗口時，如果secondv存在，顯然要更新它找出新的第二大元素.
for(size_t j=secondv+1;j<i;++j)
if(nums[j]>nums[secondv])secondv=j;
}
}
if(nums[i]>nums[maxv]){
maxv=i;
secondv=~0;
}else if(secondv==~0||nums[i]>nums[secondv]){
secondv=i;
}
out.push_back(nums[maxv]);
}
return out;
}