[Swift]LeetCode376. 摆动序列 | Wiggle Subsequence

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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 

class Solution {
    func wiggleMaxLength(_ nums: [Int]) -> Int {
        if nums.isEmpty {return 0}
        var p:Int = 1
        var q:Int = 1
        var n:Int = nums.count
        for i in 1..<n
        {
            if nums[i] > nums[i - 1]
            {
                p = q + 1
            }
            else if nums[i] < nums[i - 1]
            {
                q = p + 1
            }
        }
        return min(n, max(p, q))
    }
}

---

40ms

class Solution {
    func wiggleMaxLength(_ nums: [Int]) -> Int {

        guard nums.count > 0 else { return 0 }
        var n = nums.count
        var p = Array(repeating: 1, count: n)
        var q = Array(repeating: 1, count: n)
        for i in 1..<n {
            for j in 0..<i {
                if nums[i] < nums[j] { q[i] = max(q[i], p[j]+1)}
                else if nums[i] > nums[j] { p[i] = max(p[i], q[j]+1)}
            }
        }
        return max(q.last!, p.last!)
    }    
}