LeetCode 2765. Longest Alternating Subarray
You are given a 0-indexed integer array nums
. A subarray s
of length m
is called alternating if:
m
is greater than1
.s1 = s0 + 1
.The 0-indexed subarray
s
looks like[s0, s1, s0, s1,...,s(m-1) % 2]
. In other words,s1 - s0 = 1
,s2 - s1 = -1
,s3 - s2 = 1
,s4 - s3 = -1
, and so on up tos[m - 1] - s[m - 2] = (-1)m
.
Return the maximum length of all alternating subarrays present in nums
or -1
if no such subarray exists.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [2,3,4,3,4]
Output: 4
Explanation: The alternating subarrays are [3, 4], [3, 4, 3], and [3, 4, 3, 4]. The longest of these is [3,4,3,4], which is of length 4.
Example 2:
Input: nums = [4,5,6]
Output: 2
Explanation: [4,5] and [5,6] are the only two alternating subarrays. They are both of length 2.
Constraints:
2 <= nums.length <= 100
1 <= nums[i] <= 104
-------------------------------
解:先判断第一项是否符合,如果不符合,直接返回,然后去遍历每个元素是否跟它后面第2个数字是否一致即可。这个函数写好后,再去依次遍历,判断是否符合。下面是代码;
Runtime: 37 ms, faster than 33.33% of Java online submissions for Longest Alternating Subarray.
Memory Usage: 43.5 MB, less than 66.67% of Java online submissions for Longest Alternating Subarray.