Q:
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。
示例 1:
输入:[9,4,2,10,7,8,8,1,9] 输出:5 解释:(A[1] > A[2] < A[3] > A[4] < A[5])
示例 2:
输入:[4,8,12,16] 输出:2
示例 3:
输入:[100] 输出:1
链接:https://leetcode-cn.com/problems/longest-turbulent-subarray/description/
思路:利用动态规划的思路,设置状态矩阵dp,状态转移是根据所给条件,A[i - 1] > A[i] and A[i - 1] > A[i - 2]) or (A[i - 1] < A[i] and A[i - 1] < A[i - 2]两者之一,dp[i] = dp[i-1]+1,否则判断当前值是否与前值相同,如果不相同,dp[i] = 2,否则 dp[i]就是0
代码:
class Solution(object):
def maxTurbulenceSize(self, A):
"""
:type A: List[int]
:rtype: int
"""
n = len(A)
if n == 1:
return 1
dp = [0 for _ in range(n)]
dp[0] = 1
if A[0] == A[1]:
dp[1] = 1
else:
dp[1] = 2
for i in range(2,n):
if (A[i - 1] > A[i] and A[i - 1] > A[i - 2]) or (A[i - 1] < A[i] and A[i - 1] < A[i - 2]):
dp[i] = dp[i-1]+1
elif A[i - 1] != A[i]:
dp[i] = 2
return max(dp)