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 non-equal 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
Solution 1: Dynamic Programming
We define $f[i]$ as the length of the wiggle sequence ending at the $i$th element with an upward trend, and $g[i]$ as the length of the wiggle sequence ending at the $i$th element with a downward trend. Initially, $f[0] = g[0] = 1$ because when there is only one element, the length of the wiggle sequence is $1$. Initialize the answer as $1$.
For $f[i]$, where $i \geq 1$, we enumerate $j$ in the range $[0, i)$, if $nums[j] < nums[i]$, it means that $i$ can be appended after $j$ to form an upward wiggle sequence, then $f[i] = \max(f[i], g[j] + 1)$; if $nums[j] > nums[i]$, it means that $i$ can be appended after $j$ to form a downward wiggle sequence, then $g[i] = \max(g[i], f[j] + 1)$. Then we update the answer to $\max(f[i], g[i])$.
Finally, we return the answer.
The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
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