Leetcode|Sliding Window Maximum(multiset,優先佇列,雙端佇列和區間樹的應用)
阿新 • • 發佈:2018-11-14
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]
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]
.
1,暴力解法:最低效的時間複雜度O(k*(n-k+1));,
2,STL中的內建容器的排序功能。(O(n*logk))
為了熟悉STL,使用優先佇列和nultiset來解決,雖然時間複雜度不是最優的。
//藉助紅黑樹的排序方法,在leetcode擊敗15%的人,效率很低啦。
//藉助STL的heap操作,擊敗了20%l的leetcoder。vector<int> maxSlidingWindow(vector<int>& nums, int k) {//利用multiset中紅黑樹的排序功能 時間複雜度O(nlogk) int n=nums.size(); vector<int> max; multiset<int> smax; for(int i=0;i<n;i++){ if(i>=k) smax.erase(smax.find(nums[i-k]));//每次刪除過期的一個元素,每次都得對k個數重新排序 smax.insert(nums[i]); if(i>=k-1) max.push_back(*smax.rbegin());//找到排在最後的數,就是最大的 } return max; }
vector<int> maxSlidingWindow(vector<int>& nums, int k) {//優先佇列是一種配接器,底層是vector
int n=nums.size();
vector<int> max;
priority_queue<pair<int,int>> qmax;
for(int i=0;i<n;i++){
while(!qmax.empty()&&qmax.top().second<=i-k){
qmax.pop();
}
qmax.push(make_pair(nums[i],i));
if(i>=k-1) max.push_back(qmax.top().first);
}
return max;
}
3,線段樹(時間複雜度O(n*log(k)) 擊敗了26%的leetcoder.
線段樹節點的三個數值元素為:start,end,max(記錄區間的最大值),是以一個vector作為底層容器實現。
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
int n=nums.size();
vector<int> res;
if(k<=0||n==0||k>n) return res;
SegmentTreeNode *root=build_segmentTree(0,k-1,nums);
res.push_back(root->max);
for(int i=k;i<n;i++){
modify(i%k,nums[i],root);
res.push_back(root->max);
}
return res;
}
private:
struct SegmentTreeNode{
int start,end,max;
SegmentTreeNode *left,*right;
SegmentTreeNode(int x,int y,int m):start(x),end(y),max(m),left(NULL),right(NULL){ }
};
SegmentTreeNode* build_segmentTree(int start,int end,vector<int>& a){
if(start>end||a.size()==0||end>=a.size()) return NULL;
SegmentTreeNode* root=new SegmentTreeNode(start,end,a[start]);
if(start==end) return root;
int mid=(start+end)/2;
root->left=build_segmentTree(start,mid,a);
root->right=build_segmentTree(mid+1,end,a);
root->max=max(root->left->max,root->right->max);
return root;
}
void modify(int dot,int newval,SegmentTreeNode* root){
if(dot<root->start||dot>root->end) return;
if(root->start==root->end){
root->max=newval;
return;
}
int mid=(root->start+root->end)/2;
if(dot<=mid) modify(dot,newval,root->left);
else modify(dot,newval,root->right);
root->max=max(root->left->max,root->right->max);
}
};
4,,最優解O(n):雙端佇列 擊敗50%leetcoder
deque來儲存下標值,這樣可以獲得元素大小和位置兩組資訊。佇列元素個數永遠小於等於k,從大到小排序。擊敗50%的leetcoder
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
int n=nums.size();
vector<int> max;
deque<int> qmax;
for(int i=0;i<n;i++){
while(!qmax.empty()&&nums[qmax.back()]<=nums[i]){//隊尾的值實時更新
qmax.pop_back();
}
qmax.push_back(i);
if(qmax.front()==i-k){//更新過期的隊首
qmax.pop_front();
}
if(i>=k-1){
max.push_back(nums[qmax.front()]);
}
}
return max;
}