题目
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]
when k
is odd, and A[k] < A[k+1]
when k
is even;
- OR, for
i <= k < j
, A[k] > A[k+1]
when k
is even, and A[k] < A[k+1]
when k
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
题目大意
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
解题思路
- 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。用相邻元素差的乘积大于零(a ^ b >= 0 说明a b乘积大于零)来判断是否是湍流, 如果是,那么扩大窗口。否则窗口缩小为0,开始新的一个窗口。