376. Wiggle Subsequence

缺乏、安全感 2022-04-23 09:04 220阅读 0赞

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; 
        }
    };