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376. Wiggle Subsequence
Description
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 nonequal 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?
Solutions
DP
We maintain two dp arrays p and q,
p[i]
indicates the maximum length of the swing subsequence with a positive first difference when the position i is reachedq[i]
indicates the maximum length of the wobble subsequence with a negative first difference when reaching the i position
We traverse the array from i=1, and then for each traversed number, traverse to this number from the beginning position, then compare nums[i] and nums[j], and update the corresponding positions respectively.
Greedy
Maintain two variables p and q, and then traverse the array
 If the current number is greater than the previous number, then
p=q+1
 If the current number is smaller than the previous number, then
q=p+1
Finally, compare the larger value of p and q with n, and choose the smaller one.

class Solution { public int wiggleMaxLength(int[] nums) { int up = 1, down = 1; for (int i = 1; i < nums.length; ++i) { if (nums[i] > nums[i  1]) { up = Math.max(up, down + 1); } else if (nums[i] < nums[i  1]) { down = Math.max(down, up + 1); } } return Math.max(up, down); } }

class Solution { public: int wiggleMaxLength(vector<int>& nums) { int up = 1, down = 1; for (int i = 1; i < nums.size(); ++i) { if (nums[i] > nums[i  1]) { up = max(up, down + 1); } else if (nums[i] < nums[i  1]) { down = max(down, up + 1); } } return max(up, down); } };

class Solution: def wiggleMaxLength(self, nums: List[int]) > int: up = down = 1 for i in range(1, len(nums)): if nums[i] > nums[i  1]: up = max(up, down + 1) elif nums[i] < nums[i  1]: down = max(down, up + 1) return max(up, down) ############ class Solution(object): def wiggleMaxLength(self, nums): """ :type nums: List[int] :rtype: int """ if not nums: return 0 up = down = 1 for i in range(1, len(nums)): if nums[i] > nums[i  1]: up = down + 1 elif nums[i] < nums[i  1]: down = up + 1 return max(up, down)

func wiggleMaxLength(nums []int) int { up, down := 1, 1 for i := 1; i < len(nums); i++ { if nums[i] > nums[i1] { up = max(up, down+1) } else if nums[i] < nums[i1] { down = max(down, up+1) } } return max(up, down) }

function wiggleMaxLength(nums: number[]): number { let up = 1, down = 1; for (let i = 1; i < nums.length; ++i) { let prev = nums[i  1], cur = nums[i]; if (cur > prev) { up = Math.max(up, down + 1); } else if (cur < prev) { down = Math.max(down, up + 1); } } return Math.max(up, down); }