Leetcode 978 Longest Turbulent Subarray. (滑动窗口)

Leetcode 978

问题描述

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.

例子

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

方法一

** Solution Java **
** 4ms, 95.62% **
** 43.5MB, 62.50% **
class Solution {
    public int maxTurbulenceSize(int[] A) {
        int dec = 1, inc = 1, res = 1;
        for (int i = 1; i < A.length; ++i) {
            if (A[i - 1] < A[i]) {
                inc = dec + 1;
                dec = 1;
            } else if (A[i] < A[i - 1]){
                dec = inc + 1;
                inc = 1;
            } else {
                inc = 1;
                dec = 1;
            }
            res = Math.max(res, Math.max(dec, inc));
        }
        return res;
    }
}

方法二

** Solution Java **
** 4ms, 95.62% **
** 43.2MB, 62.50% **
class Solution {
    public int maxTurbulenceSize(int[] A) {
        int pre = 0, cur = 0, len = 1, res = 1;
        for (int i = 1; i < A.length; ++i) {
            cur = Integer.compare(A[i], A[i - 1]);
            if (cur * pre == -1) ++len;
            else if (cur == 0) len = 1;
            else len = 2;
            res = Math.max(res, len);
            pre = cur;
        }
        return res;
    }
}