376. Wiggle Subsequence

最新推荐文章于 2019-02-21 21:01:51 发布

万里归来少年心

最新推荐文章于 2019-02-21 21:01:51 发布

阅读量279

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 文章标签：贪心摇摆序列状态机 376. Wiggle Subsequence

本文链接：https://blog.youkuaiyun.com/liyazhen2011/article/details/86696596

LeetCode 专栏收录该内容

85 篇文章

订阅专栏

本文探讨了一种特殊的数列——摇摆序列，并提供了一种高效算法来寻找给定序列中最长的摇摆子序列。摇摆序列是指序列中相邻元素的差值正负交替出现的序列。通过记录序列中前后元素的状态并更新状态，可以在线性时间内找到最长的摇摆子序列。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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?

一个整数序列，如果两个相邻元素的差恰好正负交替出现，则该序列被称为摇摆序列。一个小于2个元素的序列直接为摇摆序列。给一个随机序列，求这个序列满足摇摆序列定义的最长子序列的长度。

解法：记录序列中前后两个元素的状态。初始状态为begin，如果后一个元素大于前一个元素，则状态为up，反之状态为down。当状态转换时，摇摆序列的长度加1。

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
       if (nums.size() <2)
		  return nums.size();

 	const int begin = 0;
	const int up = 1;
	const int down = -1;
	int state = begin;
	int count = 1;
	for (int i = 1; i < nums.size(); i++)
	{
		if (nums[i] > nums[i - 1])
		{
			if (state == begin || state == down)
			{
				count++;
				state = up;
			}
		}
		else if(nums[i] < nums[i - 1])
		{
			if (state == begin || state == up)
			{
				count++;
				state = down;
			}
		}
	}
	return count; 
    }
};