标签：
Given an array of n integer with duplicate number, and a moving window(size k), move the window at each iteration from the start of the array, find the maximum number inside the window at each moving.
For array [1, 2, 7, 7, 8], moving window size k = 3. return [7, 7, 8]
At first the window is at the start of the array like this
[|1, 2, 7| ,7, 8] , return the maximum 7;
then the window move one step forward.
[1, |2, 7 ,7|, 8], return the maximum 7;
then the window move one step forward again.
[1, 2, |7, 7, 8|], return the maximum 8;
o(n) time and O(k) memory
Key to the solution is to maintain a deque (size <= k) whose elements are always in descending order. Then, the first element is what we want.
Every time we want to add a new element, we need to check:
1. whether it is bigger than previous elements in deque.
If yes, we remove elements in deque which are smaller than current element.
2. whether the first element in deque is out of current sliding window.
If yes, we remove first element.
1 public class Solution { 2 3 public ArrayList<Integer> maxSlidingWindow(int[] nums, int k) { 4 ArrayList<Integer> result = new ArrayList<Integer>(); 5 Deque<Integer> deque = new LinkedList<Integer>(); 6 int i = 0; 7 for (int current : nums) { 8 i++; 9 // Ensure current deque is in decending order 10 while (!deque.isEmpty() && deque.peekLast() < current) 11 deque.pollLast(); 12 deque.addLast(current); 13 if (i > k && deque.peekFirst() == nums[i - k - 1]) 14 deque.pollFirst(); 15 if (i >= k) 16 result.add(deque.peekFirst()); 17 } 18 return result; 19 } 20 }
标签：
原文地址：http://www.cnblogs.com/ireneyanglan/p/4908033.html
