动态规划-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?

题意解读：给定一个整数序列，返回最长的wiggle sequence.wiggle sequece可以通过删除原序列的一些元素。但不打乱原序列的元素顺序

// we can use two arrays up[] and down[] to record the max wiggle sequence length so far at index i./*if(nums[i]>nums[i-1]) up[i] = down[i-1]+1 down[i] = down[i-1]if(nums[i])<nums[i-1] down[i] = up[i-1]+1 up[i] = up[i-1]if(nums[i]==nums[i-1]) down[i] = down[i-1] up[i] = up[i-1]*/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];        //初始化        down[0] = 1;up[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]){                down[i] = up[i-1]+1;                up[i] = up[i-1];            }else{                up[i] = up[i-1];                down[i] = down[i-1];            }        }        return Math.max(up[nums.length-1],down[nums.length-1]);        }}