### Problem statement

A **wiggle sequence** is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.

For example,

`[1, 7, 4, 9, 2, 5]`

is a wiggle sequence because the differences`(6, -3, 5, -7, 3)`

alternate between positive and negative.In contrast,

`[1, 4, 7, 2, 5]`

and`[1, 7, 4, 5, 5]`

are not wiggle sequences. The first is not because its first two differences are positive, and the second is not because its last difference is zero.

A **subsequence** is obtained by deleting some elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.

Given an integer array `nums`

, return the length of the longest wiggle subsequence of `nums`

.

#### Example 1

```
Input: nums = [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence with differences (6, -3, 5, -7, 3).
```

#### Example 2

```
Input: nums = [1,17,5,10,13,15,10,5,16,8]
Output: 7
Explanation: There are several subsequences that achieve this length.
One is [1, 17, 10, 13, 10, 16, 8] with differences (16, -7, 3, -3, 6, -8).
```

#### Example 3

```
Input: nums = [1,2,3,4,5,6,7,8,9]
Output: 2
```

#### Constraints

`1 <= nums.length <= 1000`

.`0 <= nums[i] <= 1000`

.

**Follow up**: Could you solve this in `O(n)`

time?

### Solution: Counting the local extrema of `nums`

First, if you pick all local extrema (minima and maxima) of `nums`

to form a subsequence `e`

, then it is wiggle. Let us call it an **extrema** subsequence.

#### Example 2

For `nums = [1,17,5,10,13,15,10,5,16,8]`

, the local extrema are `[1,17,5,15,5,16,8]`

. It is wiggle and called **extrema** subsequence.

Note that if `nums.length = n`

then `nums[0]`

and `nums[n - 1]`

are always the first and the last extremum.

Second, given any two successive local extrema `a`

and `b`

, you cannot have any wiggle subsequence between them. Because the elements between them are either monotonic increasing or decreasing.

That proves the extrema subsequence is the longest wiggle one.

#### Code

```
#include <iostream>
#include <vector>
using namespace std;
int wiggleMaxLength(vector<int>& nums) {
// nums[0] is always the first extremum
// start to find the second extremum
int i = 1;
while (i < nums.size() && nums[i] == nums[i - 1]) {
i++;
}
if (i == nums.size()) {
// all nums[i] are equal
return 1;
}
int sign = nums[i] > nums[i - 1] ? 1 : -1;
int count = 2;
i++;
while (i < nums.size()) {
if ((nums[i] - nums[i - 1]) * sign < 0) {
// nums[i] is an extremum
count++;
sign = -sign;
}
i++;
}
return count;
}
int main() {
vector<int> nums{1,7,4,9,2,5};
cout << wiggleMaxLength(nums) << endl;
nums = {1,17,5,10,13,15,10,5,16,8};
cout << wiggleMaxLength(nums) << endl;
nums = {1,2,3,4,5,6,7,8,9};
cout << wiggleMaxLength(nums) << endl;
}
```

```
Output:
6
7
2
```

#### Complexity

Runtime:

`O(n)`

, where`n = nums.length`

.Extra space:

`O(1)`

.

### References

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