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Formatted question description: https://leetcode.ca/all/376.html
376 Wiggle Subsequence
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?
@tagdp
Algorithm
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.
Code
Java

import java.util.Arrays; public class Wiggle_Subsequence { public static void main(String[] args) { Wiggle_Subsequence out = new Wiggle_Subsequence(); Solution s = out.new Solution(); System.out.println(s.wiggleMaxLength(new int[]{1,17,5,10,13,15,10,5,16,8})); } public class Solution_greedy { public int wiggleMaxLength(int[] nums) { if (nums.length < 2) { return nums.length; } int up = 1, down = 1, n = nums.length; for (int i = 1; i < n; ++i) { // start from 2nd if (nums[i] > nums[i  1]) up = down + 1; else if (nums[i] < nums[i  1]) down = up + 1; } return Math.min(n, Math.max(up, down)); } } public class Solution_better_oN { public int wiggleMaxLength(int[] nums) { if (nums.length < 2) { return nums.length; } int n = nums.length; int[] up = new int[n]; int[] down = new int[n]; up[0] = down[0] = 1; for (int i = 1; i < n; i++) { if (nums[i] > nums[i  1]) { up[i] = down[i  1] + 1; down[i] = down[i  1]; } else if (nums[i] < nums[i  1]) { down[i] = up[i  1] + 1; up[i] = up[i  1]; } else { // @note: test case, input: [0,0] down[i] = down[i  1]; up[i] = up[i  1]; } } return Math.max(down[n  1], up[n  1]); } } public class Solution { // o(N^2) // @note: could start with either up or down, both valid public int wiggleMaxLength(int[] nums) { if (nums.length < 2) { return nums.length; } // up[i] refers to the length of the longest wiggle subsequence obtained so far // considering i^th element as the last element of the wiggle subsequence and ending with a rising wiggle. int[] up = new int[nums.length]; int[] down = new int[nums.length]; Arrays.fill(up, 1); // initialization, each position is at least 1 for that char Arrays.fill(down, 1); // @note: or don't do this step, in final return just `1+()` for (int i = 1; i < nums.length; i++) { for(int j = 0; j < i; j++) { if (nums[i] > nums[j]) { up[i] = Math.max(up[i],down[j] + 1); } else if (nums[i] < nums[j]) { down[i] = Math.max(down[i],up[j] + 1); } } } return Math.max(down[nums.length  1], up[nums.length  1]); } } }

// OJ: https://leetcode.com/problems/wigglesubsequence/ // Time: O(N) // Space: O(1) class Solution { public: int wiggleMaxLength(vector<int>& nums) { int inc = 1, dec = 1, N = nums.size(); for (int i = 1; i < N; ++i) { if (nums[i] > nums[i  1]) inc = dec + 1; else if (nums[i] < nums[i  1]) dec = inc + 1; } return max(inc, dec); } };

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)