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76 lines
2.9 KiB
Markdown
76 lines
2.9 KiB
Markdown
# [376. Wiggle Subsequence](https://leetcode.com/problems/wiggle-subsequence/)
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## 题目
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Given an integer array `nums`, return *the length of the longest **wiggle sequence***.
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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 fewer than two elements is trivially a wiggle sequence.
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- 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.
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- 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.
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A **subsequence** is obtained by deleting some elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.
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**Example 1:**
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```
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Input: nums = [1,7,4,9,2,5]
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Output: 6
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Explanation: The entire sequence is a wiggle sequence.
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```
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**Example 2:**
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```
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Input: nums = [1,17,5,10,13,15,10,5,16,8]
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Output: 7
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Explanation: There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
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```
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**Example 3:**
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```
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Input: nums = [1,2,3,4,5,6,7,8,9]
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Output: 2
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```
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**Constraints:**
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- `1 <= nums.length <= 1000`
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- `0 <= nums[i] <= 1000`
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**Follow up:** Could you solve this in `O(n)` time?
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## 题目大意
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如果连续数字之间的差严格地在正数和负数之间交替,则数字序列称为摆动序列。第一个差(如果存在的话)可能是正数或负数。少于两个元素的序列也是摆动序列。例如, [1,7,4,9,2,5] 是一个摆动序列,因为差值 (6,-3,5,-7,3) 是正负交替出现的。相反, [1,4,7,2,5] 和 [1,7,4,5,5] 不是摆动序列,第一个序列是因为它的前两个差值都是正数,第二个序列是因为它的最后一个差值为零。给定一个整数序列,返回作为摆动序列的最长子序列的长度。 通过从原始序列中删除一些(也可以不删除)元素来获得子序列,剩下的元素保持其原始顺序。
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## 解题思路
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- 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。
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## 代码
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```go
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package leetcode
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func wiggleMaxLength(nums []int) int {
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if len(nums) < 2 {
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return len(nums)
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}
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res := 1
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prevDiff := nums[1] - nums[0]
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if prevDiff != 0 {
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res = 2
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}
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for i := 2; i < len(nums); i++ {
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diff := nums[i] - nums[i-1]
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if diff > 0 && prevDiff <= 0 || diff < 0 && prevDiff >= 0 {
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res++
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prevDiff = diff
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}
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}
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return res
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}
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``` |