A subarray A[i], A[i+1], ..., A[j]
of A
is said to be turbulent if and only if:
- For
i <= k < j
,A[k] > A[k+1]
whenk
is odd, andA[k] < A[k+1]
whenk
is even; - OR, for
i <= k < j
,A[k] > A[k+1]
whenk
is even, andA[k] < A[k+1]
whenk
is odd.
That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.
Return the length of a maximum size turbulent subarray of A.
Example 1:
Input: [9,4,2,10,7,8,8,1,9]
Output: 5
Explanation: (A[1] > A[2] < A[3] > A[4] < A[5])
Example 2:
Input: [4,8,12,16]
Output: 2
Example 3:
Input: [100]
Output: 1
Note:
1 <= A.length <= 40000
0 <= A[i] <= 10^9
当 A 的子数组 A[i], A[i+1], ..., A[j] 满足下列条件时,我们称其为湍流子数组:
若 i <= k < j,当 k 为奇数时, A[k] > A[k+1],且当 k 为偶数时,A[k] < A[k+1]; 或 若 i <= k < j,当 k 为偶数时,A[k] > A[k+1] ,且当 k 为奇数时, A[k] < A[k+1]。 也就是说,如果比较符号在子数组中的每个相邻元素对之间翻转,则该子数组是湍流子数组。
返回 A 的最大湍流子数组的长度。
提示:
- 1 <= A.length <= 40000
- 0 <= A[i] <= 10^9
- 给出一个数组,要求找出“摆动数组”的最大长度。所谓“摆动数组”的意思是,元素一大一小间隔的。
- 这一题可以用滑动窗口来解答。用一个变量记住下次出现的元素需要大于还是需要小于前一个元素。也可以用模拟的方法,用两个变量分别记录上升和下降数字的长度。一旦元素相等了,上升和下降数字长度都置为 1,其他时候按照上升和下降的关系增加队列长度即可,最后输出动态维护的最长长度。