[LeetCode][JavaScript]Sliding Window Maximum

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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?

Hint:

1. How about using a data structure such as deque (double-ended queue)?
2. The queue size need not be the same as the window’s size.
3. Remove redundant elements and the queue should store only elements that need to be considered

https://leetcode.com/problems/sliding-window-maximum/

 

 

 

 

 

首先是非線性複雜度的解法，直接用js的數組來類比雙向隊列。

 1 /** 2  * @param {number[]} nums 3  * @param {number} k 4  * @return {number[]} 5  */ 6 var maxSlidingWindow = function(nums, k) { 7     var window = [], result = []; 8     for(var i = 0; i < nums.length; i++){ 9         if(window.length === k){10             result.push(findMax(window));11             window.shift();12         }13         window.push(nums[i]);14     }15     if(window.length === k){16         result.push(findMax(window));17     }18     return k !== 0 ? result : [];19 20     function findMax(arr){21         var max = -Infinity;22         for(var i = 0; i < arr.length; i++){23             if(arr[i] > max){24                 max = arr[i];25             }26         }27         return max;28     }29 };

 

線性複雜度的解法。

https://leetcode.com/discuss/46578/java-o-n-solution-using-deque-with-explanationa

核心思想就是在隊列頭儲存最大元素的下標。

每一輪迴圈：

1.先移除到期的元素；

2.從後往前移出不可能是結果的元素，兩種情況：

i)比如隊列是[2,3]，這一輪的元素是4，那麼max一定是4，2和3都可以移除了；

ii)比如隊列是[5,3]，這一輪的元素是4，那麼要移出3，最後的結果應該是[5,4]。

因為等5到期了，4可能就是最大的元素了，所以要放著。為什麼可以放心地移除3呢，因為4的到期時間肯定比3久，而且大於3。 <-- 此處是痛點

3.往隊列裡放入這一輪的元素，如果長度大於等於視窗長度與就輸出。

儲存下標而非值是為了之後可以方便地找出到期的元素，需要值的時候直接可以通過下標訪問nums。

 

 1 /** 2  * @param {number[]} nums 3  * @param {number} k 4  * @return {number[]} 5  */ 6 var maxSlidingWindow = function(nums, k) { 7     var window = [], index = -1, result = []; 8     for(var i = 0; i < nums.length; i++){ 9         while(window.length !== 0 && window[0] < i - k + 1){10             window.shift();11         }12         while(window.length !== 0 && nums[window[window.length - 1]] < nums[i]){13             window.pop();14         }15         window.push(i);16         if(i >= k - 1){17             result.push(nums[window[0]]);18         }19     }20     return result;   21 };

 

 

 

 

 

[LeetCode][JavaScript]Sliding Window Maximum