Leetcode 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?

Code (Java):

class Solution {
    public int wiggleMaxLength(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int[] dp = new int[nums.length];
        int[] mode = new int[nums.length];
        
        int maxLen = 1;
        for (int i = 0; i < nums.length; i++) {
            dp[i] = 1;
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j] && mode[j] < 1) {
                    if (dp[i] < dp[j] + 1) {
                        dp[i] = dp[j] + 1;
                        mode[i] = 1;
                    }
                } else if (nums[i] < nums[j] && mode[j] > -1) {
                    if (dp[i] < dp[j] + 1) {
                        dp[i] = dp[j] + 1;
                        mode[i] = -1;
                    }
                }
            }
            
            maxLen = Math.max(maxLen, dp[i]);
        }
        
        return dp[nums.length - 1];
    }
}

O(n) time solution: