Leetcode 376. Wiggle Subsequence

-题目

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: 6The entire sequence is a wiggle sequence.Input: [1,17,5,10,13,15,10,5,16,8]Output: 7There 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

Follow up:
Can you do it in O(n) time?
翻译：
题目的意思就是给定一个序列然后求其中包含的最长  wigglee sequence的长度，wiggle sequence就是相邻两数之差正负相间的序列。

-思路

看到要求是O(n)的算法的时候，我就抛弃了dp的解法。发现这道题其实有一定的规律，只要按照从左到右的顺序，每次留下和之前的差值正负相间的元素就好。

-代码

class Solution {public:    int wiggleMaxLength(vector<int>& nums) {        if(nums.size() <= 2) {            if(nums.size() == 2 && nums[0] == nums[1]) return 1;             else return nums.size();         }        int sta = 0;         int max_len = 1;         for(int i = 1; i < nums.size(); i++) {            if(nums[i] > nums[i-1] && (sta == -1 || sta == 0)) {                max_len++; sta = 1;             }            else if(nums[i] < nums[i-1] && (sta == 1 || sta == 0)) {                max_len++; sta = -1;             }        }        return max_len;     }};