leetcode 376. Wiggle Subsequence 最长摆动序列 + 动态规划DP

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

想了好久没想到怎么做，网上看到了一个DP的做法，真的很妙。

使用用up[i]和down[i]分别记录到第i个元素为止以上升沿和下降沿结束的最长“摆动”序列长度，遍历数组，如果nums[i]>nums[i-1]，表明第i-1到第i个元素是上升的，因此up[i]只需在down[i-1]的基础上加1即可，而down[i]保持down[i-1]不变；如果nums[i]

/* * 用up[i]和down[i]分别记录到第i个元素为止以上升沿和下降沿结束的最长“摆动” * 序列长度，遍历数组，如果nums[i]>nums[i-1]，表明第i-1到第i个元素是上升的， * 因此up[i]只需在down[i-1]的基础上加1即可，而down[i]保持down[i-1]不变； * 如果nums[i]<nums[i-1]，表明第i-1到第i个元素是下降的，因此down[i] * 只需在up[i-1]的基础上加1即可，而up[i]保持up[i-1]不变；如果nums[i]==nums[i-1]， * 则up[i]保持up[i-1]，down[i]保持down[i-1]。比较最终以上升沿和下降沿结束的 * 最长“摆动”序列长度即可获取最终结果 * */class Solution {    public int wiggleMaxLength(int[] nums)     {        if(nums==null || nums.length<=0)            return 0;        int[] up=new int[nums.length];        int[] down=new int[nums.length];        up[0]=down[0]=1;        for(int i=1;i<nums.length;i++)        {            if(nums[i]>nums[i-1])            {                up[i]=down[i-1]+1;                down[i]=down[i-1];            }else if(nums[i]<nums[i-1])            {                up[i]=up[i-1];                down[i]=up[i-1]+1;            }else               {                up[i]=up[i-1];                down[i]=down[i-1];            }        }        return Math.max(up[nums.length-1], down[nums.length-1]);    }}