# 978. Longest Turbulent Subarray

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`

``````class Solution {
public:
int maxTurbulenceSize(vector<int>& arr) {
int res = 1, n = arr.size();
vector<int> dec(n, 1), inc(n, 1);
for (int i = 1; i < n; ++i) {
if (arr[i - 1] > arr[i]) {
dec[i] = inc[i - 1] + 1;
} else if (arr[i - 1] < arr[i]) {
inc[i] = dec[i - 1] + 1;
}
res = max(res, max(dec[i], inc[i]));
}
return res;
}
};
``````

``````class Solution {
public:
int maxTurbulenceSize(vector<int>& arr) {
int res = 1, n = arr.size(), inc = 1, dec = 1;
for (int i = 1; i < n; ++i) {
if (arr[i] < arr[i - 1]) {
dec = inc + 1;
inc = 1;
} else if (arr[i] > arr[i - 1]) {
inc = dec + 1;
dec = 1;
} else {
inc = 1;
dec = 1;
}
res = max(res, max(inc, dec));
}
return res;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/978

Maximum Subarray

https://leetcode.com/problems/longest-turbulent-subarray/

https://leetcode.com/problems/longest-turbulent-subarray/discuss/221935/Java-O(N)-time-O(1)-space

https://leetcode.com/problems/longest-turbulent-subarray/discuss/221929/C%2B%2BJava-6-lines-flip-the-sign

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