# 376 Wiggle Subsequence

## 376. [Wiggle Subsequence](https://leetcode.com/problems/wiggle-subsequence/description/)

## 1. Question

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.

**Examples:**

```
Input:
 [1,7,4,9,2,5]

Output:
 6
The entire sequence is a wiggle sequence.


Input:
 [1,17,5,10,13,15,10,5,16,8]

Output:
 7
There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].


Input:
 [1,2,3,4,5,6,7,8,9]

Output:
 2
```

**Follow up:**\
Can you do it in O(n) time?

## 2. Implementation

**(1) DP**

```java
class Solution {
    public int wiggleMaxLength(int[] nums) {
        if (nums.length <= 1) {
            return nums.length;
        }

        int n = nums.length;
        int[][] dp = new int[2][n];
        Arrays.fill(dp[0], 1);
        Arrays.fill(dp[1], 1);
        int maxLen = 1;

        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i] && dp[1][j] + 1 > dp[0][i]) {
                    dp[0][i] = dp[1][j] + 1;
                }
                else if (nums[j] > nums[i] && dp[0][j] + 1 > dp[1][i]) {
                    dp[1][i] = dp[0][j] + 1;
                }
            }
            maxLen = Math.max(maxLen, Math.max(dp[0][i], dp[1][i]));
        }
        return maxLen;
    }
}
```

## 3. Time & Space Complexity

**(1) DP:** 时间复杂度O(n^2), 空间复杂度O(n)
