Given an integer array nums, return the length of the longest wiggle sequence.

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 fewer than two elements is trivially a wiggle sequence.

• For example, [1, 7, 4, 9, 2, 5] is a wiggle sequence because the differences (6, -3, 5, -7, 3) are alternately positive and negative.
• In contrast, [1, 4, 7, 2, 5] and [1, 7, 4, 5, 5] are not wiggle sequences, the first because its first two differences are positive and the second because its last difference is zero.

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

Example 1:

Input: nums = [1,7,4,9,2,5] Output: 6 Explanation: The entire sequence is a wiggle sequence. 

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]. 

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?

如果连续数字之间的差严格地在正数和负数之间交替，则数字序列称为摆动序列。第一个差（如果存在的话）可能是正数或负数。少于两个元素的序列也是摆动序列。例如， [1,7,4,9,2,5] 是一个摆动序列，因为差值 (6,-3,5,-7,3) 是正负交替出现的。相反, [1,4,7,2,5] 和 [1,7,4,5,5] 不是摆动序列，第一个序列是因为它的前两个差值都是正数，第二个序列是因为它的最后一个差值为零。给定一个整数序列，返回作为摆动序列的最长子序列的长度。 通过从原始序列中删除一些（也可以不删除）元素来获得子序列，剩下的元素保持其原始顺序。

• 题目要求找到摆动序列最长的子序列。本题可以用贪心的思路，记录当前序列的上升和下降的趋势。扫描数组过程中，每扫描一个元素都判断是“峰”还是“谷”，根据前一个是“峰”还是“谷”做出对应的决定。利用贪心的思想找到最长的摆动子序列。

package leetcode funcwiggleMaxLength(nums []int) int { iflen(nums) <2 { returnlen(nums) } res:=1prevDiff:=nums[1] -nums[0] ifprevDiff!=0 { res=2 } fori:=2; i<len(nums); i++ { diff:=nums[i] -nums[i-1] ifdiff>0&&prevDiff<=0||diff<0&&prevDiff>=0 { res++prevDiff=diff } } returnres }