<DQueue、双端队列、滑动窗口>239.Sliding Window Maximum

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大学里的混子

修改于 2019-02-25 11:00:11

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修改于 2019-02-25 11:00:11

文章被收录于专栏：LeetCode

一、239.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. Return the max sliding window.

Example:

Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7] 
Explanation: 

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

时间复杂度是O(N)

 
  public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null ||k < 1||nums.length<k) return new int[nums.length];
        int [] res = new int[nums.length - k + 1];
        LinkedList<Integer> DQueue = new LinkedList<>();
        int index = 0;
        for (int i = 0 ; i < nums.length;i++){
            //当前的树与队列尾部的数相比较，如果尾部的数比当前的数小，则弹出
            while (!DQueue.isEmpty() && nums[DQueue.getLast()] <= nums[i]){
                DQueue.pollLast();
            }
            DQueue.addLast(i);
            
            //加入了新的数，那么就要将头部的数删除
            if (DQueue.getFirst() == i - k){
                DQueue.pollFirst();
            }
            if (i >= k - 1){
                 res[index++] = nums[DQueue.peekFirst()];
            }
        }
        return res;
    }

二、求子数组中的最大值与最小值之差小于num的个数

O(N)的解法，用一个双端队列记录窗口中最大值，一个记录最小值。

public static int allLessNumSubArr(int [] arr,int num){
    if (arr == null || arr.length == 0) return 0;
    int res = 0;
    LinkedList <Integer> max = new LinkedList<>();
    LinkedList <Integer> min = new LinkedList<>();
    int right = 0;
    for (int left = 0 ;left < arr.length; left++){
        while(right < arr.length){
            while (!max.isEmpty() && arr[max.peekLast()] <= arr[right]){
                max.pollLast();
            }
            max.addLast(right);
            while (!min.isEmpty() && arr[min.peekLast()] >= arr[right]){
                min.pollLast();
            }
            min.addLast(right);

            if (arr[max.peekFirst()] - arr[min.peekFirst()] > num)
                break;
            right++;
        }
        if (min.peekFirst() == left){
            min.pollFirst();
        }
        if (max.peekFirst() == left){
            max.pollFirst();
        }

        res += right - left;
        left++;
    }
    return res;
}

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