leetcode-376. Wiggle Subsequence_a sequence of n numbers is given.a prefix of a seq-优快云博客

本文介绍了一种求解最长摆动子序列的算法。该算法首先去除原序列中的重复元素，然后通过判断相邻元素差值的正负来确定是否为摆动序列，并计算其最大长度。

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题目：

链接：https://leetcode.com/problems/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.

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
题目的意思很简单，在数组中找到一个最长的子序列，使得这个序列的任意三个连续数字之间的差是相反的。也就是子序列中的数字的差是正负(或是负正交替出现的)。

源码：

class Solution {
public:

	bool is_same(int pre,int next){
		if (pre > 0 && next > 0)return false;
		else if (pre <0 && next <0)return false;
		else return true;
	}

	int wiggleMaxLength(vector<int>& nums) {
		vector<int>num;
		int size = nums.size();
		for (int i = 0; i < size; i++){
			if(num.size()==0)num.push_back(nums[i]);
			else{
				if (nums[i] != *--num.end())num.push_back(nums[i]);
			}
		}
		int size2 = num.size();
		if (size2 == 0)return 0;
		if (size2 == 1)return 1;
		vector<int>diff(size2 - 1, 0);
		vector<int>re(size2-1,0);
		for (int i = 1; i < size2; i++)diff[i - 1] = num[i] - num[i - 1];
		re[0] = 1;
		for (int i = 1; i < size2 - 1; i++){
			if (is_same(diff[i - 1],diff[i]))re[i] = re[i - 1] + 1;
			else re[i] = re[i - 1];
		}
		return *--re.end() + 1;
	}
};