• Sliding Window Maximum


    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, ie: 1 ≤ k ≤ input array's size for non-empty array.

    Follow up:
    Could you solve it in linear time?

    Analyse: Find the maximum in each window and push it into the result vector.

    Runtime: 340ms.

     1 class Solution {
     2 public:
     3     vector<int> maxSlidingWindow(vector<int>& nums, int k) {
     4         vector<int> result;
     5         if(nums.size() == 0) return result;
     6         
     7         int left = 0, right = k - 1;
     8         while(right < nums.size()){
     9             int temp = INT_MIN; //maximum in the current window
    10             if(left == right) temp = nums[left];
    11             else{
    12                 for(int i = left; i <= right; i++)
    13                     temp = max(temp, nums[i]);
    14             }
    15             result.push_back(temp);
    16             left++;
    17             right++;
    18         }
    19         return result;
    20     }
    21 };
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  • 原文地址:https://www.cnblogs.com/amazingzoe/p/4799783.html
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