A sequence of numbers is called a wiggle sequence if 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.
Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.
Example 1:
Input: [1,7,4,9,2,5] Output: 6 Explanation: The entire sequence is a wiggle sequence.
Example 2:
Input: [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: [1,2,3,4,5,6,7,8,9] Output: 2
Follow up:
Can you do it 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。
示例 1:
输入: [1,7,4,9,2,5] 输出: 6 解释: 整个序列均为摆动序列。
示例 2:
输入: [1,17,5,10,13,15,10,5,16,8] 输出: 7 解释: 这个序列包含几个长度为 7 摆动序列，其中一个可为[1,17,10,13,10,16,8]。
示例 3:
输入: [1,2,3,4,5,6,7,8,9] 输出: 2
进阶:
你能否用 O(n) 时间复杂度完成此题?
12ms
1 class Solution {
2 func wiggleMaxLength(_ nums: [Int]) -> Int {
3 if nums.isEmpty {return 0}
4 var p:Int = 1
5 var q:Int = 1
6 var n:Int = nums.count
7 for i in 1..<n
8 {
9 if nums[i] > nums[i - 1]
10 {
11 p = q + 1
12 }
13 else if nums[i] < nums[i - 1]
14 {
15 q = p + 1
16 }
17 }
18 return min(n, max(p, q))
19 }
20 }
40ms
1 class Solution {
2 func wiggleMaxLength(_ nums: [Int]) -> Int {
3
4 guard nums.count > 0 else { return 0 }
5 var n = nums.count
6 var p = Array(repeating: 1, count: n)
7 var q = Array(repeating: 1, count: n)
8 for i in 1..<n {
9 for j in 0..<i {
10 if nums[i] < nums[j] { q[i] = max(q[i], p[j]+1)}
11 else if nums[i] > nums[j] { p[i] = max(p[i], q[j]+1)}
12 }
13 }
14 return max(q.last!, p.last!)
15 }
16 }
原文地址：https://www.cnblogs.com/strengthen/p/10278265.html