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]whenkis odd, andarr[k] < arr[k + 1]whenkis even.
- Or, for
i <= k < j:arr[k] > arr[k + 1]whenkis even, andarr[k] < arr[k + 1]whenkis 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 * 1040 <= arr[i] <= 109
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
}