978. Longest Turbulent Subarray

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. 1 <= A.length <= 40000
2. 0 <= A[i] <= 10^9

class Solution {
      public int maxTurbulenceSize(int[] A) {
            if (A.length == 0) return 0;

            int n = A.length, maxLen = 0;
            int[][] state = new int[n][2];

            for (int i = 1; i < n; i++) {
                if (A[i - 1] < A[i]) {
                    state[i][0] = state[i - 1][1] + 1;
                    maxLen = Math.max(maxLen, state[i][0]);
                } else if (A[i - 1] > A[i]) {
                    state[i][1] = state[i - 1][0] + 1;  
                    maxLen = Math.max(maxLen, state[i][1]);
                }
            }

            return maxLen + 1;
        }
}

 啥意思？反正turbulent只要保证大小大或者小大小就行了，那如果当前的比前一个大，就更新以当前为结尾且当前比前一个大的dp值就可以，反之亦然

用当前比前一个大举例，i > i -1, 要更新他，他是从i-1 < i - 2得来的，就用dp[i - 1][1] + 1和1中较大值。每次更新