LeetCode 978. Longest Turbulent Subarray

原题链接在这里：https://leetcode.com/problems/longest-turbulent-subarray/

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

题解：

Set some small examples like [1, 3, 2], [2,2] and find routine.

It matters the last 3 componenets. If it is l<m>r or l>m<r relationship, then length+1. Otherwise, reset to 2 or 1.

Let dp[i] denotes up to A[i-1], the longest turbulent length.

If  A[i-3]<A[i-2]>A[i-1] or  A[i-3]>A[i-2]<A[i-1], dp[i] = dp[i-1] + 1.

Maintain the maximum to res. 

Note: corner case {1, 1, 2}. The final res needs to be updated again. Otherwise, the result is 1 not 2.

Time Complexity: O(n). n = A.length.

Space: O(n).

AC Java:

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

        if(A.length < 2){
            return A.length;
        }

        int len = A.length;
        int [] dp = new int[len+1];
        dp[1] = 1;
        dp[2] = A[0] == A[1] ? 1 : 2;

        int res = dp[2];
        for(int i = 3; i<=len; i++){
            if(A[i-2]<A[i-3] && A[i-2]<A[i-1] || A[i-2]>A[i-3] && A[i-2]>A[i-1]){
                dp[i] = dp[i-1] + 1;
                res = Math.max(res, dp[i]);
            }else if(A[i-1] == A[i-2]){
                dp[i] = 1;
            }else{
                dp[i] = 2;
                res = Math.max(res, dp[i]);
            }
        }

        return res;
    }
}

It only cares about dp[i-1]. Thus it could reduce dimension.

Time Complexity: O(n).

Space: O(1).

AC Java:

class Solution {
    public int maxTurbulenceSize(int[] arr) {
        if(arr == null){
            return 0;
        }

        if(arr.length < 2){
            return arr.length;
        }

        int n = arr.length;
        int dp = arr[0] == arr[1] ? 1 : 2;
        int res = dp;
        for(int i = 3; i <= n; i++){
            if((arr[i - 3] < arr[i - 2] && arr[i - 1] < arr[i - 2]) || (arr[i - 3] > arr[i - 2] && arr[i - 2] < arr[i - 1])){
                dp = dp + 1;
                res = Math.max(res, dp);
            }else if(arr[i - 1] == arr[i - 2]){
                dp = 1;
            }else{
                dp = 2;
            }
        }

        res = Math.max(res, dp);
        return res;
    }
}

类似Maximum Subarray.