114 lines
3.2 KiB
C++
114 lines
3.2 KiB
C++
// Source : https://leetcode.com/problems/sliding-window-maximum/
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// Author : Hao Chen
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// Date : 2015-07-19
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/**********************************************************************************
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*
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* Given an array nums, there is a sliding window of size k which is moving from the
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* very left of the array to the very right. You can only see the k numbers in the
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* window. Each time the sliding window moves right by one position.
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*
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* For example,
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* Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
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*
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* Window position Max
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* --------------- -----
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* [1 3 -1] -3 5 3 6 7 3
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* 1 [3 -1 -3] 5 3 6 7 3
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* 1 3 [-1 -3 5] 3 6 7 5
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* 1 3 -1 [-3 5 3] 6 7 5
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* 1 3 -1 -3 [5 3 6] 7 6
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* 1 3 -1 -3 5 [3 6 7] 7
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*
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* Therefore, return the max sliding window as [3,3,5,5,6,7].
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*
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* Note:
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* You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty
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* array.
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*
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* Follow up:
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* Could you solve it in linear time?
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*
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* How about using a data structure such as deque (double-ended queue)?
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* The queue size need not be the same as the window’s size.
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* Remove redundant elements and the queue should store only elements that need to be
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* considered.
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*
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**********************************************************************************/
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#include <iostream>
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#include <vector>
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#include <deque>
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#include <set>
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using namespace std;
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//O(nlog(k)
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vector<int> maxSlidingWindow02(vector<int>& nums, int k) {
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vector<int> result;
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//using multiset for collecting the window data (O(nlog(k) time complexity)
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multiset<int> w;
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for(int i=0; i<nums.size(); i++) {
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//remove the left item which leaves window
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if (i >= k) {
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w.erase(w.find(nums[i-k]));
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}
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//insert the right itme which enter the window
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w.insert(nums[i]);
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if (i>=k-1) {
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result.push_back(*w.rbegin());
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}
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}
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return result;
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}
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//O(n)
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vector<int> maxSlidingWindow01(vector<int>& nums, int k) {
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vector<int> result;
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//using multiset for collecting the window data (O(nlog(k) time complexity)
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deque<int> q;
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for(int i=0; i<nums.size(); i++) {
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//remove the left item which leaves window
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if (!q.empty() && q.front() == i - k) {
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q.pop_front();
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}
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//remove all num which less than current number from the back one by one
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while (!q.empty() && nums[q.back()] < nums[i]) {
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q.pop_back();
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}
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//insert the right itme which enter the window
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q.push_back(i);
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if (i>=k-1) {
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result.push_back(nums[q.front()]);
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}
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}
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return result;
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}
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vector<int> maxSlidingWindow(vector<int>& nums, int k) {
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return maxSlidingWindow01(nums, k);
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return maxSlidingWindow02(nums, k);
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}
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void printVector( vector<int>& v ) {
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cout << "{ ";
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for(int i=0; i<v.size(); i++) {
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cout << v[i] << (i==v.size() ? " ": ", ");
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}
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cout << "}" << endl;
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}
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int main(int argc, char** argv)
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{
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int a[] = {1,3,-1,-3,5,3,6,7};
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int k = 3;
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vector<int> nums(a, a+sizeof(a)/sizeof(a[0]));
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printVector(nums);
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vector<int> result = maxSlidingWindow(nums, k);
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printVector(result);
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}
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