zoukankan      html  css  js  c++  java
  • Sliding Window Maximum 解答

    Question

    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.

    Example

    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;

    Challenge

    o(n) time and O(k) memory

    Solution

    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 }
  • 相关阅读:
    LeetCode
    <OFFER15> 15_NumberOf1InBinary
    《OFFER14》14_CuttingRope
    Convert DataFrame string complex i to j python // “Cloning” row or column vectors
    Sorting arrays in NumPy by column
    【説明する】深度优先及宽度优先算法比较
    codevs 2924 数独挑战 x(三种做法+超详细注释~)
    第四次考试大整理
    洛谷 P1048 采药
    codevs 3137-3139 栈练习 x
  • 原文地址:https://www.cnblogs.com/ireneyanglan/p/4908033.html
Copyright © 2011-2022 走看看