978. Longest Turbulent Subarray

Medium

Given an integer array arr, return the length of a maximum size turbulent subarray of arr.

A subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.

More formally, a subarray [arr[i], arr[i + 1], ..., arr[j]] of arr is said to be turbulent if and only if:

• For i <= k < j:

  • arr[k] > arr[k + 1] when k is odd, and

  • arr[k] < arr[k + 1] when k is even.

• Or, for i <= k < j:

  • arr[k] > arr[k + 1] when k is even, and

  • arr[k] < arr[k + 1] when k is odd.

Example 1:

Input: arr = [9,4,2,10,7,8,8,1,9]
Output:
 5
Explanation:
 arr[1] > arr[2] < arr[3] > arr[4] < arr[5]

Example 2:

Input: arr = [4,8,12,16]
Output:
 2

Example 3:

Input: arr = [100]
Output:
 1

Constraints:

• 1 <= arr.length <= 4 * 104

• 0 <= arr[i] <= 109

解題

func maxTurbulenceSize(arr []int) int {
    if len(arr)==1 { return 1 }
    
    count := 0
    ans := 0
    ascending := true
    
    for i:=0; i<len(arr)-1; i++ {
        if arr[i]>arr[i+1] { 
            ascending= false 
            count++
            break
        } else if arr[i]<arr[i+1] { 
            ascending= true
            count++
            break
        }  
    }
    
    for i:=1; i<len(arr)-1; i++ {
        if arr[i]< arr[i+1] && ascending==false {
            fmt.Println(arr[i], arr[i+1])
            count++
            ascending = true
        } else if arr[i]> arr[i+1] && ascending==true {
            fmt.Println(arr[i], arr[i+1])
            count++ 
            ascending = false
        } else {
            if count > ans { ans = count }
            
            count = 0
            if arr[i]>arr[i+1] { 
                ascending= false 
                count++
            }
            if arr[i]<arr[i+1] { 
                ascending= true
                count++
            }
            
        }
    }
    
    if count > ans { ans = count }
    
    return ans+1
}