0376. Wiggle Subsequence

# 376. Wiggle Subsequence#

## 题目 #

Given an integer array `nums`, return the length of the longest wiggle sequence.

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 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.

subsequence is obtained by deleting some elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.

Example 1:

``````Input: nums = [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence.
``````

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].
``````

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?

## 解题思路 #

• 题目要求找到摆动序列最长的子序列。本题可以用贪心的思路，记录当前序列的上升和下降的趋势。扫描数组过程中，每扫描一个元素都判断是“峰”还是“谷”，根据前一个是“峰”还是“谷”做出对应的决定。利用贪心的思想找到最长的摆动子序列。

## 代码 #

``````package leetcode

func wiggleMaxLength(nums []int) int {
if len(nums) < 2 {
return len(nums)
}
res := 1
prevDiff := nums - nums
if prevDiff != 0 {
res = 2
}
for i := 2; i < len(nums); i++ {
diff := nums[i] - nums[i-1]
if diff > 0 && prevDiff <= 0 || diff < 0 && prevDiff >= 0 {
res++
prevDiff = diff
}
}
return res
}
`````` Jun 22, 2022 Edit this page